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<?xml version="1.0" encoding="utf-8"?>
<search>
<entry>
<title></title>
<url>/2023/04/09/C++%E8%B0%83%E7%94%A8Python3.7/</url>
<content><![CDATA[<p>@<a href="C++%E8%B0%83%E7%94%A8Python3.7">TOC</a></p>
<h1 id="前言"><a href="#前言" class="headerlink" title="前言"></a>前言</h1><p>最近在做一个项目，涉及到了如何用C#调用Python文件。汇总了一下网上找来的资料，具体参考了这篇博客<a href="https://blog.csdn.net/qq_42063091/article/details/82418630">c#调用python的四种方法（尝试了四种，只详细讲解本人成功的后两种，其余方法只列出，详细用法请自行谷歌百度）</a>。<br>对于第一种方法，使用IronPython2.7的方法，我找到了更详细的博客<a href="https://blog.csdn.net/cw19901024/article/details/73526402">C#如何调用Python执行脚本，并将执行结果显示值显示至C#界面</a>，不过Python2.7可能无法满足我们项目的需求，当然我也找到了类似的另外两种可以使用Python3的方法，还没有试过，在这里提供给大家参考<a href="https://blog.csdn.net/Micusd/article/details/81605593">C#使用公共语言拓展（CLE）调用Python3（使用TensorFlow训练的模型）</a>、<a href="https://blog.csdn.net/weixin_42388228/article/details/89681870">pythonnet c#调用并集成python代码</a>。<br>对于第三和第四种方法，因为要使用命令行窗口，而且还要考虑不同用户的python安装情况如何，在项目实际使用时效果肯定是很不好的，所以在这里不做过多的介绍，原博主也给出了他自己的源代码，大家可以自己尝试一下。<br>重点说一下第二种方法。我在CSDN上找了好多博客，大家的方法都大同小异，可能真的是看人品吧，我试了好几次都失败了，最后在朋友的帮助下才成功……</p>
<h1 id="配置环境"><a href="#配置环境" class="headerlink" title="配置环境"></a>配置环境</h1><p>这里先说一下，我是win10+VS2017+Python3.7.4。</p>
<h2 id="第一步新建一个控制台项目，修改属性页"><a href="#第一步新建一个控制台项目，修改属性页" class="headerlink" title="第一步新建一个控制台项目，修改属性页"></a>第一步新建一个控制台项目，修改属性页</h2><p>选择Debug|x64<br><img src="https://img-blog.csdnimg.cn/2019081916113616.JPG?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzQxMjE0NjEw,size_16,color_FFFFFF,t_70" alt="在这里插入图片描述">第一个修改的是C/C++——常规——附加包含目录：添加你所安装的Python目录下的include目录路径。<br><img src="https://img-blog.csdnimg.cn/201908191614311.JPG?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzQxMjE0NjEw,size_16,color_FFFFFF,t_70" alt="在这里插入图片描述">第二个修改的是链接器——常规——附加库目录：添加你所安装的Python目录下的libs目录路径。</p>
<h2 id="第二步修改Python安装目录下文件"><a href="#第二步修改Python安装目录下文件" class="headerlink" title="第二步修改Python安装目录下文件"></a>第二步修改Python安装目录下文件</h2><p><img src="https://img-blog.csdnimg.cn/20190819161713384.JPG?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzQxMjE0NjEw,size_16,color_FFFFFF,t_70" alt="在这里插入图片描述"><br>找到Python安装目录，找到libs目录复制python37.lib到本目录，改名为python37_d.lib。</p>
<h1 id="代码测试"><a href="#代码测试" class="headerlink" title="代码测试"></a>代码测试</h1><p>现在我们可以来简单测试一下</p>
<h2 id="测试用Python3语法输出“Hello-World！”"><a href="#测试用Python3语法输出“Hello-World！”" class="headerlink" title="测试用Python3语法输出“Hello World！”"></a>测试用Python3语法输出“Hello World！”</h2><p>我们定义一个print()，里面用了Python3的语法。</p>
<figure class="highlight c"><table><tr><td class="code"><pre><span class="line"><span class="meta">#<span class="keyword">include</span> <span class="string">"pch.h"</span></span></span><br><span class="line"><span class="meta">#<span class="keyword">include</span> <span class="string"><Python.h></span></span></span><br><span class="line"><span class="meta">#<span class="keyword">include</span> <span class="string"><iostream></span></span></span><br><span class="line"></span><br><span class="line">using namespace <span class="built_in">std</span>;</span><br><span class="line"></span><br><span class="line"><span class="type">void</span> <span class="title function_">print</span><span class="params">()</span></span><br><span class="line">{</span><br><span class="line"> Py_Initialize();</span><br><span class="line"> PyRun_SimpleString(<span class="string">"print('Hello World!')\n"</span>);<span class="comment">//Python3语法</span></span><br><span class="line"> Py_Finalize();</span><br><span class="line">}</span><br><span class="line"></span><br><span class="line"><span class="type">int</span> <span class="title function_">main</span><span class="params">()</span></span><br><span class="line">{</span><br><span class="line"> print();</span><br><span class="line"> <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
<h2 id="测试用Python文件输出“Hello-World！”"><a href="#测试用Python文件输出“Hello-World！”" class="headerlink" title="测试用Python文件输出“Hello World！”"></a>测试用Python文件输出“Hello World！”</h2><p>定义一个Hello()，里面调用了Test001.py文件。</p>
<figure class="highlight c"><table><tr><td class="code"><pre><span class="line"><span class="meta">#<span class="keyword">include</span> <span class="string">"pch.h"</span></span></span><br><span class="line"><span class="meta">#<span class="keyword">include</span> <span class="string"><Python.h></span></span></span><br><span class="line"><span class="meta">#<span class="keyword">include</span> <span class="string"><iostream></span></span></span><br><span class="line"></span><br><span class="line">using namespace <span class="built_in">std</span>;</span><br><span class="line"></span><br><span class="line"><span class="type">void</span> <span class="title function_">print</span><span class="params">()</span></span><br><span class="line">{</span><br><span class="line"> Py_Initialize();</span><br><span class="line"> PyRun_SimpleString(<span class="string">"print('Hello Python!')\n"</span>);</span><br><span class="line"> Py_Finalize();</span><br><span class="line">}</span><br><span class="line"></span><br><span class="line"><span class="type">void</span> <span class="title function_">Hello</span><span class="params">()</span></span><br><span class="line">{</span><br><span class="line"> Py_Initialize();<span class="comment">//调用Py_Initialize()进行初始化</span></span><br><span class="line"> PyObject * pModule = <span class="literal">NULL</span>;</span><br><span class="line"> PyObject * pFunc = <span class="literal">NULL</span>;</span><br><span class="line"> pModule = PyImport_ImportModule(<span class="string">"Test001"</span>);<span class="comment">//调用的Python文件名</span></span><br><span class="line"> pFunc = PyObject_GetAttrString(pModule, <span class="string">"Hello"</span>);<span class="comment">//调用的函数名</span></span><br><span class="line"> PyEval_CallObject(pFunc, <span class="literal">NULL</span>);<span class="comment">//调用函数,NULL表示参数为空</span></span><br><span class="line"> Py_Finalize();<span class="comment">//调用Py_Finalize,和Py_Initialize相对应的.</span></span><br><span class="line">}</span><br><span class="line"></span><br><span class="line"><span class="type">int</span> <span class="title function_">main</span><span class="params">()</span></span><br><span class="line">{</span><br><span class="line"> print();</span><br><span class="line"> Hello();</span><br><span class="line"> <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
<p>Test001.py文件内容如下：</p>
<figure class="highlight python"><table><tr><td class="code"><pre><span class="line"><span class="keyword">def</span> <span class="title function_">Hello</span>():</span><br><span class="line"> <span class="built_in">print</span>(<span class="string">"Hello World!"</span>)</span><br></pre></td></tr></table></figure>
<p>注意，要把Test001.py文件放到ConsoleApplication1.exe同级目录<br><img src="https://img-blog.csdnimg.cn/2019081916354339.JPG?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzQxMjE0NjEw,size_16,color_FFFFFF,t_70" alt="在这里插入图片描述"><br><img src="https://img-blog.csdnimg.cn/20190819163559667.JPG?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzQxMjE0NjEw,size_16,color_FFFFFF,t_70" alt="在这里插入图片描述"></p>
<h1 id="结尾"><a href="#结尾" class="headerlink" title="结尾"></a>结尾</h1><p>如果一切顺利的话，那你现在已经在控制台看到输出了；如果不顺利的话，可以找找其他的博客，或者换一个方法，不要拘泥于这一个思路……<br>从今天开始，这个博客会把我之前做项目遇到的一些问题的解决办法总结一下，也算是一种积累吧，估计等我项目做完，会写成一本《扑街实录》。</p>
<h1 id="引用"><a href="#引用" class="headerlink" title="引用"></a>引用</h1><p><a href="https://blog.csdn.net/qq_42063091/article/details/82418630">c#调用python的四种方法（尝试了四种，只详细讲解本人成功的后两种，其余方法只列出，详细用法请自行谷歌百度）</a><br><a href="https://blog.csdn.net/cw19901024/article/details/73526402">C#如何调用Python执行脚本，并将执行结果显示值显示至C#界面</a><br><a href="https://blog.csdn.net/Micusd/article/details/81605593">C#使用公共语言拓展（CLE）调用Python3（使用TensorFlow训练的模型）</a><br><a href="https://blog.csdn.net/weixin_42388228/article/details/89681870">pythonnet c#调用并集成python代码</a></p>
]]></content>
</entry>
<entry>
<title>C++ vector基本操作</title>
<url>/2023/04/09/C++%20vector%E6%9F%A5%E6%94%B9%E5%A2%9E%E5%88%A0%E6%93%8D%E4%BD%9C/</url>
<content><![CDATA[<h1 id="初始化"><a href="#初始化" class="headerlink" title="初始化"></a>初始化</h1><p>在c++中，vector是一个类模板，当使用模板的时候，我们需要指出编译器应该把类和函数实例化成何种类型。</p>
<figure class="highlight cpp"><table><tr><td class="code"><pre><span class="line">vector<<span class="type">int</span>> ivec;<span class="comment">//vector的元素是int型数据</span></span><br><span class="line">vector<vector<string>> file;<span class="comment">//vector的元素还是是vector对象，这个vector对象的元素是string型数据</span></span><br></pre></td></tr></table></figure>
<span id="more"></span>
<h2 id="默认初始化"><a href="#默认初始化" class="headerlink" title="默认初始化"></a>默认初始化</h2><figure class="highlight cpp"><table><tr><td class="code"><pre><span class="line">vector<T> v1;<span class="comment">//v1是一个空vector，它潜在的元素是T类型的，执行默认初始化</span></span><br></pre></td></tr></table></figure>
<h2 id="拷贝初始化"><a href="#拷贝初始化" class="headerlink" title="拷贝初始化"></a>拷贝初始化</h2><figure class="highlight cpp"><table><tr><td class="code"><pre><span class="line"><span class="comment">//针对的是变量</span></span><br><span class="line"><span class="function">vector<T> <span class="title">v2</span><span class="params">(v1)</span></span>;<span class="comment">//v2中包含v1所有元素的副本</span></span><br><span class="line">vector<T> v2=v1;<span class="comment">//等价于v2(v1),v2中包含有v1所有元素的副本</span></span><br></pre></td></tr></table></figure>
<h2 id="列表初始化（花括号）"><a href="#列表初始化（花括号）" class="headerlink" title="列表初始化（花括号）"></a>列表初始化（花括号）</h2><figure class="highlight cpp"><table><tr><td class="code"><pre><span class="line"><span class="comment">//针对的是变量的元素</span></span><br><span class="line">vector<T> v3{a,b,c...};<span class="comment">//v3包含了初始值个数的元素，每个元素被赋予相应的初始值</span></span><br><span class="line">vector<T> v3={a,b,c};<span class="comment">//等价于上一个</span></span><br></pre></td></tr></table></figure>
<h2 id="值初始化（小括号）"><a href="#值初始化（小括号）" class="headerlink" title="值初始化（小括号）"></a>值初始化（小括号）</h2><figure class="highlight cpp"><table><tr><td class="code"><pre><span class="line"><span class="comment">//针对的是变量的元素数量</span></span><br><span class="line"><span class="comment">//输入的n应该是给构造函数的</span></span><br><span class="line"><span class="function">vector<T> <span class="title">v4</span><span class="params">(n,val)</span></span>;<span class="comment">//v5中包含着n个重复的，值为val的元素</span></span><br><span class="line"><span class="function">vector<T> <span class="title">v5</span><span class="params">(n)</span></span>;<span class="comment">//v5包含了n个重复的，默认值的元素</span></span><br></pre></td></tr></table></figure>
<p><strong>注意，使用花括号进行值初始化时，编译器也能理解</strong></p>
<figure class="highlight cpp"><table><tr><td class="code"><pre><span class="line">vector<string> v6{<span class="string">"hi"</span>};<span class="comment">//列表初始化：v6有一个元素</span></span><br><span class="line"><span class="function">vector<string> <span class="title">v7</span><span class="params">(<span class="string">"hi"</span>)</span></span>;<span class="comment">//错误，不可以用元素值来构建vector对象</span></span><br><span class="line">vector<string> v8{<span class="number">10</span>};<span class="comment">//v8有10个默认初始化的元素</span></span><br><span class="line">vector<string> v9{<span class="number">10</span>,<span class="string">"hi"</span>};<span class="comment">//v9有10个值为hi的元素</span></span><br></pre></td></tr></table></figure>
<h1 id="查找元素"><a href="#查找元素" class="headerlink" title="查找元素"></a>查找元素</h1><p>我们可以通过遍历vector，返回所查找元素的下标。</p>
<figure class="highlight cpp"><table><tr><td class="code"><pre><span class="line"><span class="function"><span class="type">int</span> <span class="title">findElement</span><span class="params">(vector<<span class="type">int</span>> v, <span class="type">int</span> key)</span></span>{</span><br><span class="line"> <span class="comment">// 获取vector的长度。</span></span><br><span class="line"> <span class="type">int</span> len = v.<span class="built_in">size</span>();</span><br><span class="line"> <span class="keyword">for</span>(<span class="type">int</span> i = <span class="number">0</span>; i < len; i++){</span><br><span class="line"> <span class="keyword">if</span>(v[i] == key){</span><br><span class="line"> <span class="keyword">return</span> i;</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> <span class="number">-1</span>;</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
<h1 id="修改元素"><a href="#修改元素" class="headerlink" title="修改元素"></a>修改元素</h1><p>直接对应下标指向位置进行修改。</p>
<figure class="highlight cpp"><table><tr><td class="code"><pre><span class="line">v[<span class="number">1</span>] = v[<span class="number">0</span>];</span><br></pre></td></tr></table></figure>
<h1 id="增加元素"><a href="#增加元素" class="headerlink" title="增加元素"></a>增加元素</h1><p>在vector中增加元素包括两种，一种是在尾部增加元素，另一种是在指定位置增加元素。</p>
<h2 id="尾部增加"><a href="#尾部增加" class="headerlink" title="尾部增加"></a>尾部增加</h2><figure class="highlight cpp"><table><tr><td class="code"><pre><span class="line"><span class="comment">// 使用push_back()函数</span></span><br><span class="line">vector<<span class="type">int</span>> v;</span><br><span class="line"><span class="keyword">for</span> (<span class="type">int</span> i = <span class="number">0</span>; i < <span class="number">10</span>; i++)</span><br><span class="line">{</span><br><span class="line"> v.<span class="built_in">push_back</span>(i);</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
<h2 id="指定位置增加"><a href="#指定位置增加" class="headerlink" title="指定位置增加"></a>指定位置增加</h2><figure class="highlight cpp"><table><tr><td class="code"><pre><span class="line"><span class="comment">// 使用insert()函数</span></span><br><span class="line"><span class="comment">// 在v[2]位置插入10</span></span><br><span class="line">v.<span class="built_in">insert</span>(v.<span class="built_in">begin</span>() + <span class="number">2</span>, <span class="number">10</span>);</span><br></pre></td></tr></table></figure>
<h1 id="删除元素"><a href="#删除元素" class="headerlink" title="删除元素"></a>删除元素</h1><p>在vector中删除元素包括三种，第一种是删除尾部的元素，第二种是删除指定的元素，第三种是删除所有元素。</p>
<h2 id="尾部删除"><a href="#尾部删除" class="headerlink" title="尾部删除"></a>尾部删除</h2><figure class="highlight cpp"><table><tr><td class="code"><pre><span class="line"><span class="comment">// 使用pop_back()函数</span></span><br><span class="line">v.<span class="built_in">pop_back</span>();</span><br></pre></td></tr></table></figure>
<h2 id="指定位置删除"><a href="#指定位置删除" class="headerlink" title="指定位置删除"></a>指定位置删除</h2><figure class="highlight cpp"><table><tr><td class="code"><pre><span class="line"><span class="comment">// 使用erase()函数</span></span><br><span class="line"><span class="comment">// 删除开始位置的元素，并不会回收空间</span></span><br><span class="line">v.<span class="built_in">erase</span>(v.<span class="built_in">begin</span>());</span><br><span class="line"><span class="comment">// 删除区间[i,j-1]的元素</span></span><br><span class="line">v.<span class="built_in">erase</span>(v.<span class="built_in">begin</span>() + i, v.<span class="built_in">end</span>() - j);</span><br></pre></td></tr></table></figure>
<h2 id="删除所有元素"><a href="#删除所有元素" class="headerlink" title="删除所有元素"></a>删除所有元素</h2><figure class="highlight cpp"><table><tr><td class="code"><pre><span class="line"><span class="comment">// 使用clear()函数</span></span><br><span class="line"><span class="comment">// 并不会回收空间，但v.size()变成0</span></span><br><span class="line">v.<span class="built_in">clear</span>();</span><br></pre></td></tr></table></figure>
]]></content>
<categories>
<category>计算机技术</category>
</categories>
</entry>
<entry>
<title></title>
<url>/2023/04/09/ILSVRC2012%E4%B8%8B%E8%BD%BD+%E8%AE%AD%E7%BB%83/</url>
<content><![CDATA[<p>@<a href="ILSVRC2012%E4%B8%8B%E8%BD%BD+%E8%AE%AD%E7%BB%83">TOC</a></p>
<h1 id="下载"><a href="#下载" class="headerlink" title="下载"></a>下载</h1><h2 id="官网下载"><a href="#官网下载" class="headerlink" title="官网下载"></a>官网下载</h2><p>直接在<a href="http://www.image-net.org/challenges/LSVRC/2012/downloads">ILSVRC2012</a>官网进行下载，需要注册账号登陆。<br><strong>训练集</strong><br>训练集下载地址：<a href="http://www.image-net.org/challenges/LSVRC/2012/dd31405981ef5f776aa17412e1f0c112/ILSVRC2012_img_train.tar">http://www.image-net.org/challenges/LSVRC/2012/dd31405981ef5f776aa17412e1f0c112/ILSVRC2012_img_train.tar</a><br><strong>验证集</strong><br>验证集下载地址：<a href="http://www.image-net.org/challenges/LSVRC/2012/dd31405981ef5f776aa17412e1f0c112/ILSVRC2012_img_val.tar">http://www.image-net.org/challenges/LSVRC/2012/dd31405981ef5f776aa17412e1f0c112/ILSVRC2012_img_val.tar</a></p>
<h2 id="迅雷下载"><a href="#迅雷下载" class="headerlink" title="迅雷下载"></a>迅雷下载</h2><p><strong>训练集</strong><br>训练集种子：<a href="http://academictorrents.com/download/a306397ccf9c2ead27155983c254227c0fd938e2.torrent">http://academictorrents.com/download/a306397ccf9c2ead27155983c254227c0fd938e2.torrent</a><br><strong>验证集</strong><br>验证集种子：<a href="http://academictorrents.com/download/5d6d0df7ed81efd49ca99ea4737e0ae5e3a5f2e5.torrent">http://academictorrents.com/download/5d6d0df7ed81efd49ca99ea4737e0ae5e3a5f2e5.torren</a></p>
<h2 id="使用aria2加速下载"><a href="#使用aria2加速下载" class="headerlink" title="使用aria2加速下载"></a>使用aria2加速下载</h2><figure class="highlight bash"><table><tr><td class="code"><pre><span class="line">aria2c -x 16 -s 16 <span class="string">'http://academictorrents.com/download/5d6d0df7ed81efd49ca99ea4737e0ae5e3a5f2e5.torren'</span> <span class="string">'http://academictorrents.com/download/a306397ccf9c2ead27155983c254227c0fd938e2.torrent'</span></span><br></pre></td></tr></table></figure>
<h2 id="数据集校验"><a href="#数据集校验" class="headerlink" title="数据集校验"></a>数据集校验</h2><figure class="highlight bash"><table><tr><td class="code"><pre><span class="line"><span class="built_in">md5sum</span> ILSVRC2012_img_val.tar ILSVRC2012_img_train.tar</span><br><span class="line">29b22e2961454d5413ddabcf34fc5622 ILSVRC2012_img_val.tar</span><br><span class="line">1d675b47d978889d74fa0da5fadfb00e ILSVRC2012_img_train.tar</span><br></pre></td></tr></table></figure>
<h1 id="训练"><a href="#训练" class="headerlink" title="训练"></a>训练</h1><h2 id="解压训练集"><a href="#解压训练集" class="headerlink" title="解压训练集"></a>解压训练集</h2><p>将ILSVRC2012_img_train.tar解压，1000个类别的*.tar包。</p>
<figure class="highlight bash"><table><tr><td class="code"><pre><span class="line"><span class="built_in">mkdir</span> train</span><br><span class="line">tar -xvf ILSVRC2012_img_train.tar -C train</span><br></pre></td></tr></table></figure>
<p>然后可以使用下面这段python代码，将训练集的1000个.tar包解压缩，并删除源.tar包。</p>
<figure class="highlight python"><table><tr><td class="code"><pre><span class="line"><span class="keyword">import</span> glob</span><br><span class="line"><span class="keyword">import</span> os</span><br><span class="line"></span><br><span class="line">filelist = glob.glob(<span class="string">'./train/*.tar'</span>)</span><br><span class="line"></span><br><span class="line"><span class="keyword">for</span> f <span class="keyword">in</span> filelist:</span><br><span class="line"> os.system(<span class="string">"mkdir ./train/"</span> + f.split(<span class="string">'.'</span>)[<span class="number">0</span>])</span><br><span class="line"></span><br><span class="line"><span class="keyword">for</span> f <span class="keyword">in</span> filelist:</span><br><span class="line"> os.system(<span class="string">"tar -xvf "</span> + f + <span class="string">" -C ./train/"</span> + f.split(<span class="string">'.'</span>)[<span class="number">0</span>])</span><br><span class="line"></span><br><span class="line"><span class="keyword">for</span> f <span class="keyword">in</span> filelist:</span><br><span class="line"> os.system(<span class="string">"rm ./train/"</span> + f)</span><br></pre></td></tr></table></figure>
<h2 id="解压验证集"><a href="#解压验证集" class="headerlink" title="解压验证集"></a>解压验证集</h2><p>将ILSVRC2012_img_val.tar解压，得到没有种类标签的图片。</p>
<figure class="highlight bash"><table><tr><td class="code"><pre><span class="line"><span class="built_in">mkdir</span> val</span><br><span class="line">tar -xvf ILSVRC2012_img_val.tar -C val</span><br></pre></td></tr></table></figure>
<p>然后使用<a href="https://raw.githubusercontent.com/soumith/imagenetloader.torch/master/valprep.sh">valprep.sh</a>文件，将验证集整理为和训练集相同的格式，按照种类标签划分文件夹。</p>
<figure class="highlight bash"><table><tr><td class="code"><pre><span class="line"><span class="built_in">cd</span> val</span><br><span class="line">aria2c -x 16 -s 16 ‘https://raw.githubusercontent.com/soumith/imagenetloader.torch/master/valprep.sh’</span><br><span class="line">sh valprep.sh</span><br></pre></td></tr></table></figure>
]]></content>
</entry>
<entry>
<title></title>
<url>/2023/04/09/Jetson%20Nano%20&%20TX2%E9%85%8D%E7%BD%AE%E6%95%99%E7%A8%8B/</url>
<content><<br>更新于2021年4月10日。本文介绍了Jetson Nano 和 Jetson TX2两种设备的配置教程，前一部分以Jetson Nano为例，后半部分以Jetson TX2为例。这两种设备的配置教程，大体上是相近的。</p>
<h1 id="Jetson-Nano-激活"><a href="#Jetson-Nano-激活" class="headerlink" title="Jetson Nano 激活"></a>Jetson Nano 激活</h1><p>首先，拿到一张64GB/128GB（12GB的卡不够用）的micro-SD card，并进行格式化。SD卡的格式化工具，直接从<a href="https://www.sdcard.org/chs/downloads/formatter/index.html">官网</a>下载即可，使用时选择“快速格式化”。</p>
<p>然后，从<a href="https://developer.nvidia.com/zh-cn/embedded/jetpack#install">Nvidia官网</a>下载你需要的jetson nano镜像。</p>
<p>注意确定安装的jetpack版本，这关系到其自带的cuda版本。简单来说，jetpack4.4以上，只能安装pytorch1.6以上<a href="https://forums.developer.nvidia.com/t/pytorch-for-jetson-version-1-8-0-now-available/72048">PyTorch for Jetson - version 1.8.0 now available</a>。而且在pytorch1.6环境下保存的模型，没办法在更低版本的pytorch中打开。</p>
<p>旧版本可以从官网<a href="https://developer.nvidia.com/embedded/jetpack-archive">JetPack Archive</a>寻找。<br>最后，根据<a href="https://developer.nvidia.com/embedded/learn/get-started-jetson-nano-devkit">Nvidia官网教程</a>完成镜像的烧录，大概需要半个小时的时间。</p>
<h1 id="Jetson-TX2-烧录"><a href="#Jetson-TX2-烧录" class="headerlink" title="Jetson TX2 烧录"></a>Jetson TX2 烧录</h1><p>这里给出两个链接，一个是官网链接<a href="https://docs.nvidia.com/sdk-manager/install-with-sdkm-jetson/index.html">使用SDK Manager安装Jetson软件</a>，另一个是知乎上的链接<a href="https://zhuanlan.zhihu.com/p/94071423">Jetson TX2 入门教程（镜像烧写）</a>。</p>
<h1 id="更新系统国内源"><a href="#更新系统国内源" class="headerlink" title="更新系统国内源"></a>更新系统国内源</h1><p>在使用系统之前，我们先将系统更换为国内安装源。<br>首先，备份source.list。</p>
<figure class="highlight powershell"><table><tr><td class="code"><pre><span class="line">sudo <span class="built_in">cp</span> /etc/apt/sources.list /etc/apt/sources.list.bak</span><br></pre></td></tr></table></figure>
<p>然后，修改source.list。推荐用gedit，十分方便。如果没有安装的话，需要先安装一下。</p>
<figure class="highlight powershell"><table><tr><td class="code"><pre><span class="line">sudo apt<span class="literal">-get</span> install gedit</span><br><span class="line">sudo gedit /etc/apt/sources.list</span><br></pre></td></tr></table></figure>
<p>接着，替换list中的内容。</p>
<figure class="highlight powershell"><table><tr><td class="code"><pre><span class="line">deb http://mirrors.tuna.tsinghua.edu.cn/ubuntu<span class="literal">-ports</span>/ bionic main multiverse restricted universe</span><br><span class="line">deb http://mirrors.tuna.tsinghua.edu.cn/ubuntu<span class="literal">-ports</span>/ bionic<span class="literal">-security</span> main multiverse restricted universe</span><br><span class="line">deb http://mirrors.tuna.tsinghua.edu.cn/ubuntu<span class="literal">-ports</span>/ bionic<span class="literal">-updates</span> main multiverse restricted universe</span><br><span class="line">deb http://mirrors.tuna.tsinghua.edu.cn/ubuntu<span class="literal">-ports</span>/ bionic<span class="literal">-backports</span> main multiverse restricted universe</span><br><span class="line">deb<span class="literal">-src</span> http://mirrors.tuna.tsinghua.edu.cn/ubuntu<span class="literal">-ports</span>/ bionic main multiverse restricted universe</span><br><span class="line">deb<span class="literal">-src</span> http://mirrors.tuna.tsinghua.edu.cn/ubuntu<span class="literal">-ports</span>/ bionic<span class="literal">-security</span> main multiverse restricted universe</span><br><span class="line">deb<span class="literal">-src</span> http://mirrors.tuna.tsinghua.edu.cn/ubuntu<span class="literal">-ports</span>/ bionic<span class="literal">-updates</span> main multiverse restricted universe</span><br><span class="line">deb<span class="literal">-src</span> http://mirrors.tuna.tsinghua.edu.cn/ubuntu<span class="literal">-ports</span>/ bionic<span class="literal">-backports</span> main multiverse restricted universe</span><br></pre></td></tr></table></figure>
<p>最后，更新软件。update和upgrade的区别可以参考<a href="https://www.cnblogs.com/fenglongyu/p/8654991.html">linux命令系列 sudo apt-get update和upgrade的区别</a>。<br>这一步可能需要花一段时间。</p>
<figure class="highlight powershell"><table><tr><td class="code"><pre><span class="line">sudo apt<span class="literal">-get</span> update </span><br><span class="line">sudo apt<span class="literal">-get</span> upgrade </span><br></pre></td></tr></table></figure>
<h1 id="安装系统工具jtop"><a href="#安装系统工具jtop" class="headerlink" title="安装系统工具jtop"></a>安装系统工具jtop</h1><p>Jetson Nano中有个工具jtop, 可以查看CPU和GPU资源，还可以显示系统的jetpack版本，十分好用。<br>首先，安装pip3.</p>
<figure class="highlight powershell"><table><tr><td class="code"><pre><span class="line">sudo apt<span class="literal">-get</span> install python3<span class="literal">-pip</span></span><br></pre></td></tr></table></figure>
<p>同样的，我们也可以将pip修改为国内源。<br>第一种方式，长期修改。我以阿里源为例，其他国内源只需要修改链接即可。</p>
<figure class="highlight bash"><table><tr><td class="code"><pre><span class="line">pip3 config <span class="built_in">set</span> global.index-url https://mirrors.aliyun.com/pypi/simple/</span><br></pre></td></tr></table></figure>
<p>然后，使用pip安装jetson-stats。<br>第二种方式，一次性修改。我以清华源为例，其他国内源只需要修改链接即可。</p>
<figure class="highlight powershell"><table><tr><td class="code"><pre><span class="line">sudo pip3 install jetson<span class="literal">-stats</span> <span class="literal">-i</span> https://pypi.tuna.tsinghua.edu.cn/simple</span><br></pre></td></tr></table></figure>
<p>安装好之后，我们来看一下效果如何。</p>
<figure class="highlight powershell"><table><tr><td class="code"><pre><span class="line">sudo jtop</span><br></pre></td></tr></table></figure>
<h1 id="从源代码编译opencv"><a href="#从源代码编译opencv" class="headerlink" title="从源代码编译opencv"></a>从源代码编译opencv</h1><p>刷机后的Nano已经预装了opencv，但是预装版本不支持CUDA，所以我们还要在Jetson Nano上手动安装OpenCV。</p>
<p><strong>方法一</strong><br>安装已编译的opencv，简单方便，但是这个版本过低，且不能指定其他版本。</p>
<figure class="highlight bash"><table><tr><td class="code"><pre><span class="line">sudo apt-get install opencv-python</span><br></pre></td></tr></table></figure>
<p><strong>方法二</strong><br>从源代码安装，但是使用别人写好的脚本，如<a href="https://github.com/jetsonhacks/buildOpenCVXavier">buildOpenCVXavier</a>，<a href="https://github.com/mdegans/nano_build_opencv">nano_build_opencv</a>。但是报错之后，可能不知道自己为什么错了，因为脚本不是你自己写的。</p>
<p><strong>方法三</strong><br>自己从源代码编译安装，有很多坑要踩啊。<br>不能先安装archconda等虚拟环境管理包，否则会有问题！参考<a href="https://forums.developer.nvidia.com/t/install-opencv-for-python3-in-jetson-nano/74042">在Jetson Nano中为python3安装OpenCV</a></p>
<ol>
<li>增加交换空间</li>
</ol>
<p>opencv的编译时生成的中间文件很大，Jetson Nano原始空间放不下。这里介绍一种永久生效的方法。要是只想临时生效，可以参考<a href="https://qengineering.eu/install-opencv-4.5-on-jetson-nano.html">Enlarge memory swap</a>。</p>
<p>查看当前系统的交换空间。</p>
<figure class="highlight bash"><table><tr><td class="code"><pre><span class="line">sudo swapon --show</span><br></pre></td></tr></table></figure>
<p>查看内存大小。</p>
<figure class="highlight bash"><table><tr><td class="code"><pre><span class="line">free -h</span><br></pre></td></tr></table></figure>
<p>创建用于swap的文件。</p>
<figure class="highlight bash"><table><tr><td class="code"><pre><span class="line">sudo fallocate -l 2G /swapfile</span><br></pre></td></tr></table></figure>
<p>设置交换空间。</p>
<figure class="highlight bash"><table><tr><td class="code"><pre><span class="line">sudo mkswap /swapfile</span><br></pre></td></tr></table></figure>
<p>激活交换空间。</p>
<figure class="highlight bash"><table><tr><td class="code"><pre><span class="line">sudo swapon /swapfile</span><br><span class="line"><span class="comment">#为了使这个激活永久有效</span></span><br><span class="line">sudo gedit /etc/fstab</span><br><span class="line"><span class="comment">#粘贴 /swapfile swap swap defaults 0 0</span></span><br></pre></td></tr></table></figure>
<p>验证增加空间是否有效。</p>
<figure class="highlight bash"><table><tr><td class="code"><pre><span class="line">sudo swapon --show</span><br><span class="line">sudo free -h</span><br></pre></td></tr></table></figure>
<ol start="2">
<li>安装依赖项<br> 这个依赖项的要求五花八门，取决于你对功能的要求，这里我只给出我使用的依赖项，参考<a href="https://pythops.com/post/compile-deeplearning-libraries-for-jetson-nano">compile deeplearning libraries for jetson nano</a>。如果遇到包冲突问题，参考<a href="https://www.cnblogs.com/klcf0220/p/10242810.html">aptitude与apt-get</a>解决。</li>
</ol>
<figure class="highlight bash"><table><tr><td class="code"><pre><span class="line">dependencies=(build-essential</span><br><span class="line"> cmake</span><br><span class="line"> pkg-config</span><br><span class="line"> libavcodec-dev</span><br><span class="line"> libavformat-dev</span><br><span class="line"> libopenblas-base</span><br><span class="line"> libopenmpi-dev</span><br><span class="line"> libswscale-dev</span><br><span class="line"> libv4l-dev</span><br><span class="line"> libxvidcore-dev</span><br><span class="line"> libavresample-dev</span><br><span class="line"> python3-dev</span><br><span class="line"> python3-numpy</span><br><span class="line"> libtbb2</span><br><span class="line"> libtbb-dev</span><br><span class="line"> libtiff-dev</span><br><span class="line"> libjpeg-dev</span><br><span class="line"> libpng-dev</span><br><span class="line"> libtiff-dev</span><br><span class="line"> libdc1394-22-dev</span><br><span class="line"> libgtk-3-dev</span><br><span class="line"> libcanberra-gtk3-module</span><br><span class="line"> libatlas-base-dev</span><br><span class="line"> gfortran</span><br><span class="line"> wget</span><br><span class="line"> unzip)</span><br><span class="line"></span><br><span class="line">sudo apt install -y <span class="variable">${dependencies[@]}</span></span><br></pre></td></tr></table></figure>
<ol start="3">
<li>下载源代码<br>安装所有第三方依赖项后，我们就需要下载OpenCV啦。需要两个压缩包：基本库和其他拓展库。在<a href="https://opencv.org/releases/">opencv官网</a>下载你需要的版本。这里以4.4.0版本为例。</li>
</ol>
<p>在github网址下载opencv-4.4.0.zip，复制下面的网址，在浏览器直接下载。</p>
<figure class="highlight bash"><table><tr><td class="code"><pre><span class="line">https://github.com/opencv/opencv/archive/4.4.0.zip</span><br></pre></td></tr></table></figure>
<p>在github网址下载opencv_contrib-4.4.0.zip，复制下面的网址，在浏览器直接下载。</p>
<figure class="highlight bash"><table><tr><td class="code"><pre><span class="line">https://github.com/opencv/opencv_contrib/archive/4.4.0.zip</span><br></pre></td></tr></table></figure>
<p>对下载后的opencv-4.4.0.zip和opencv_contrib-4.4.0.zip进行解压缩。<br>在opencv-4.4.0文件夹下打开命令行，新建一个build文件夹，并切换到build文件夹下。</p>
<figure class="highlight bash"><table><tr><td class="code"><pre><span class="line"><span class="built_in">mkdir</span> build</span><br><span class="line"><span class="built_in">cd</span> build</span><br></pre></td></tr></table></figure>
<ol start="4">
<li>编写cmake配置文件my_cmake.sh<br>这里是对opencv和cuda做的一些配置，和依赖库一样也是五花八门，看你的需求吧。这里我给出自己的配置，仅供参考。</li>
</ol>
<p>新建脚本文件。</p>
<figure class="highlight bash"><table><tr><td class="code"><pre><span class="line">gedit my_cmake.sh</span><br></pre></td></tr></table></figure>
<p>粘贴cmake命令。</p>
<figure class="highlight bash"><table><tr><td class="code"><pre><span class="line">cmake -D CMAKE_BUILD_TYPE=RELEASE \</span><br><span class="line"> -D WITH_CUDA=ON \</span><br><span class="line"> -D CUDA_ARCH_PTX=<span class="string">""</span> \</span><br><span class="line"> -D CUDA_ARCH_BIN=<span class="string">"5.3,6.2,7.2"</span> \</span><br><span class="line"> -D WITH_CUBLAS=ON \</span><br><span class="line"> -D WITH_LIBV4L=ON \</span><br><span class="line"> -D BUILD_opencv_python3=ON \</span><br><span class="line"> -D BUILD_opencv_python2=OFF \</span><br><span class="line"> -D BUILD_opencv_java=OFF \</span><br><span class="line"> -D WITH_GSTREAMER=OFF \</span><br><span class="line"> -D WITH_GTK=ON \</span><br><span class="line"> -D BUILD_TESTS=OFF \</span><br><span class="line"> -D BUILD_PERF_TESTS=OFF \</span><br><span class="line"> -D BUILD_EXAMPLES=OFF \</span><br><span class="line"> -D OPENCV_EXTRA_MODULES_PATH=../../opencv_contrib-4.4.0/modules \</span><br><span class="line"> ..</span><br></pre></td></tr></table></figure>
<p>具体每个选项的介绍，可以参考<a href="https://blog.csdn.net/weixin_40784980/article/details/105246509">Jetson Nano 从头配置OpenCV+CUDA+QT完整流程</a>。</p>
<p>执行make命令。</p>
<figure class="highlight bash"><table><tr><td class="code"><pre><span class="line">sh ./my_cmake.sh</span><br></pre></td></tr></table></figure>
<p>在我们进入实际编译步骤之前，请确保检查CMake的输出！参考<a href="https://blog.csdn.net/qq_27971677/article/details/90400118">树莓派安装OpenCV-4.1.0及Contrib</a></p>
<ol start="5">
<li>开始构建<figure class="highlight bash"><table><tr><td class="code"><pre><span class="line">$ make -j4</span><br></pre></td></tr></table></figure>
-j4表示4个CPU核心同时运行。<br>查看CPU核心数，参考<a href="https://www.cnblogs.com/chenzhen0530/p/12109868.html">Ubuntu 18.04配置OpenCV 4.2.0</a>。</li>
</ol>
<figure class="highlight bash"><table><tr><td class="code"><pre><span class="line"><span class="built_in">nproc</span></span><br></pre></td></tr></table></figure>
<p>经过漫长的等待，约两小时。<br>可能遇到错误，这是网络问题导致部分依赖包没有下载。<br><strong>错误一</strong><br>参考<a href="https://blog.csdn.net/weixin_40784980/article/details/105246509">Jetson Nano 从头配置OpenCV+CUDA+QT完整流程</a>中相关部分的解决方案。</p>
<figure class="highlight bash"><table><tr><td class="code"><pre><span class="line">fatal error: boostdesc_bgm.i: No such file or directory</span><br></pre></td></tr></table></figure>
<p><strong>错误二</strong></p>
<figure class="highlight bash"><table><tr><td class="code"><pre><span class="line">fatal error: opencv2/xfeatures2d/cuda.hpp: No such file or directory</span><br></pre></td></tr></table></figure>
<p>参考<a href="https://blog.csdn.net/hongge_smile/article/details/104372783?utm_medium=distribute.pc_relevant_bbs_down.none-task-blog-baidujs-2.nonecase&depth_1-utm_source=distribute.pc_relevant_bbs_down.none-task-blog-baidujs-2.nonecase">opencv安装opencv_contrib出现无法打开包括文件: “opencv2/xfeatures2d/cuda.hpp”</a>中的解决方案。记得加在最开始，不要加在最后面！</p>
<p><strong>错误三</strong></p>
<figure class="highlight bash"><table><tr><td class="code"><pre><span class="line">CMakeFiles/example_gpu_surf_keypoint_matcher.dir/surf_keypoint_matcher.cpp.o: In <span class="keyword">function</span> `main<span class="string">':</span></span><br><span class="line"><span class="string">surf_keypoint_matcher.cpp:(.text.startup.main+0x352): undefined reference to `cv::cuda::SURF_CUDA::SURF_CUDA()'</span></span><br><span class="line">surf_keypoint_matcher.cpp:(.text.startup.main+0x579): undefined reference to `cv::cuda::SURF_CUDA::operator()(cv::cuda::GpuMat const&, cv::cuda::GpuMat const&, cv::cuda::GpuMat&, cv::cuda::GpuMat&, bool)<span class="string">'</span></span><br><span class="line"><span class="string">surf_keypoint_matcher.cpp:(.text.startup.main+0x60d): undefined reference to `cv::cuda::SURF_CUDA::operator()(cv::cuda::GpuMat const&, cv::cuda::GpuMat const&, cv::cuda::GpuMat&, cv::cuda::GpuMat&, bool)'</span></span><br><span class="line">surf_keypoint_matcher.cpp:(.text.startup.main+0x6af): undefined reference to `cv::cuda::SURF_CUDA::defaultNorm() const<span class="string">'</span></span><br><span class="line"><span class="string">surf_keypoint_matcher.cpp:(.text.startup.main+0x7ca): undefined reference to `cv::cuda::SURF_CUDA::downloadKeypoints(cv::cuda::GpuMat const&, std::vector<cv::KeyPoint, std::allocator<cv::KeyPoint> >&)'</span></span><br><span class="line">surf_keypoint_matcher.cpp:(.text.startup.main+0x7ea): undefined reference to `cv::cuda::SURF_CUDA::downloadKeypoints(cv::cuda::GpuMat const&, std::vector<cv::KeyPoint, std::allocator<cv::KeyPoint> >&)<span class="string">'</span></span><br><span class="line"><span class="string">surf_keypoint_matcher.cpp:(.text.startup.main+0x800): undefined reference to `cv::cuda::SURF_CUDA::downloadDescriptors(cv::cuda::GpuMat const&, std::vector<float, std::allocator<float> >&)'</span></span><br><span class="line">surf_keypoint_matcher.cpp:(.text.startup.main+0x812): undefined reference to `cv::cuda::SURF_CUDA::downloadDescriptors(cv::cuda::GpuMat const&, std::vector<<span class="built_in">float</span>, std::allocator<<span class="built_in">float</span>> >&)<span class="string">'</span></span><br><span class="line"><span class="string">collect2: error: ld returned 1 exit status</span></span><br><span class="line"><span class="string">samples/gpu/CMakeFiles/example_gpu_surf_keypoint_matcher.dir/build.make:132: recipe for target '</span>bin/example_gpu_surf_keypoint_matcher<span class="string">' failed</span></span><br><span class="line"><span class="string">make[2]: *** [bin/example_gpu_surf_keypoint_matcher] Error 1</span></span><br></pre></td></tr></table></figure>
<p>参考<a href="https://blog.csdn.net/GungnirsPledge/article/details/108597474#5CMakeFilesexample_gpu_surf_keypoint_matcherdirsurf_keypoint_matchercppo_In_function_main_252">源码编译安装OpenCV4.5+ CUDA11 带python 遇到的错误总结</a>中的解决方案。</p>
<ol start="6">
<li>安装</li>
</ol>
<figure class="highlight bash"><table><tr><td class="code"><pre><span class="line">make install</span><br></pre></td></tr></table></figure>
<ol start="7">
<li>验证</li>
</ol>
<figure class="highlight bash"><table><tr><td class="code"><pre><span class="line">$ python3 -c <span class="string">"import cv2; print(cv2.__version__)"</span></span><br><span class="line">4.4.0</span><br></pre></td></tr></table></figure>
<h1 id="安装archconda"><a href="#安装archconda" class="headerlink" title="安装archconda"></a>安装archconda</h1><p>anaconda是深度学习的好工具。在arm64框架平台上，对应有一个archconda，是大神编译好的工具，我们可以直接使用。<br>从大神<a href="https://github.com/Archiconda/build-tools/releases">github网址</a>上下载sh文件，然后就像在普通linux上安装即可。</p>
<p>如果需要在虚拟环境中使用opencv，有两种方法。<br><strong>方法一</strong><br>链接，参考<a href="https://www.bojankomazec.com/2019/12/installing-opencv-4-on-nvidia-jetson.html">Installing OpenCV 4 on NVIDIA Jetson Nano</a>。</p>
<figure class="highlight bash"><table><tr><td class="code"><pre><span class="line"><span class="comment"># Go to the folder where OpenCV's native library is built</span></span><br><span class="line"><span class="built_in">cd</span> /usr/local/lib/python3.6/site-packages/cv2/python-3.6</span><br><span class="line"><span class="comment"># Rename</span></span><br><span class="line"><span class="built_in">mv</span> cv2.cpython-36m-xxx-linux-gnu.so cv2.so</span><br><span class="line"><span class="comment"># Go to your virtual environments site-packages folder if previously set</span></span><br><span class="line"><span class="built_in">cd</span> ~/env/lib/python3.6/site-packages/</span><br><span class="line"><span class="comment"># Or just go to your home folder if not set a venv site-packages folder</span></span><br><span class="line"><span class="built_in">cd</span> ~</span><br><span class="line"><span class="comment"># Symlink the native library</span></span><br><span class="line"><span class="built_in">ln</span> -s /usr/local/lib/python3.6/site-packages/cv2/python-3.6/cv2.so cv2.so</span><br></pre></td></tr></table></figure>
<p><strong>方法二</strong><br>拷贝，参考<a href="https://github.com/mdegans/nano_build_opencv/issues/32">How to check for successful install including in virtualenv. #32</a>。</p>
<figure class="highlight bash"><table><tr><td class="code"><pre><span class="line"><span class="comment"># Go to the folder where OpenCV's native library is built</span></span><br><span class="line"><span class="built_in">cd</span> /usr/local/lib/python3.6/site-packages/cv2/python-3.6</span><br><span class="line"><span class="comment"># copy</span></span><br><span class="line"><span class="built_in">cp</span> usr/local/lib/python3.6/site-packages/cv2/python-3.6 [path to venv]/lib/python3.6/site-packages/cv2/python-3.6</span><br></pre></td></tr></table></figure>
<p>在虚拟环境中导入cv2，可能会遇到下面一个错误。</p>
<figure class="highlight bash"><table><tr><td class="code"><pre><span class="line">from .cv2 import *</span><br><span class="line">ImportError: numpy.core.multiarray failed to import</span><br></pre></td></tr></table></figure>
<p>这主要是在编译opencv的时候使用的numpy，和虚拟环境中安装的numpy版本不一致，根据安装时的numpy版本修改一下即可，参考<a href="https://zhuanlan.zhihu.com/p/280702247">cv2: numpy.core.multiarray failed to import</a>。<br>查看编译opencv的时候使用的numpy版本，可以进入系统自带的python环境，输出numpy版本号。</p>
<figure class="highlight bash"><table><tr><td class="code"><pre><span class="line">>>> import numpy</span><br><span class="line">>>> <span class="built_in">print</span>(numpy.__version__)</span><br></pre></td></tr></table></figure>
<p>还有可能遇到另一个问题。</p>
<figure class="highlight bash"><table><tr><td class="code"><pre><span class="line">>>> import numpy</span><br><span class="line">>>> Illegal instruction (core dumped)</span><br></pre></td></tr></table></figure>
<p>这是numpy版本的问题，将numpy=1.19.5降级到numpy=1.19.4或更低版本即可，参考<a href="https://forums.developer.nvidia.com/t/illegal-instruction-core-dumped/165488">Illegal instruction (core dumped)</a></p>
<h1 id="安装mmdetection"><a href="#安装mmdetection" class="headerlink" title="安装mmdetection"></a>安装mmdetection</h1><p>这一部分使用的是Jetson TX2，参考<a href="https://blog.csdn.net/qq_24282081/article/details/107285072#STEP1%20%E5%AE%89%E8%A3%85pytorch%C2%A0">在Jetson tx2安装 mmdetection环境</a>。</p>
<p>这里我遇到了一个小问题，TX2接上网线但是连不上网，解决方法参考<a href="https://blog.csdn.net/tao_fuqiang/article/details/79741317">Jetson TX2 有线网络网线不识别,灯不亮问题</a>。</p>
<p>其实，安装完opencv后，要运行mmdetection的代码，还有4个重要的包需要安装：torch，torchvision，mmcv-full，mmdet。</p>
<ol>
<li>安装torch</li>
</ol>
<p>去NVIDIA官网下载配套的版本<a href="https://forums.developer.nvidia.com/t/pytorch-for-jetson-version-1-7-0-now-available/72048">PyTorch for Jetson - version 1.7.0 now available</a>，这一步需要翻墙。<br>注意查看自己的Jetpack版本号，4.4要用torch1.6，否则会报错cudnn版本不合适。<br>查看Jetson设备的信息可以用之前提到的jtop，也可以使用<a href="http://www.gpus.cn/gpus_list_page_techno_support_content?id=39">查询Jetson设备与开发环境版本的基础信息</a>。<br>下载好pytorch之后，激活虚拟环境，使用pip命令安装。</p>
<figure class="highlight bash"><table><tr><td class="code"><pre><span class="line">pip install torch-1.6.0-cp36-cp36m-linux_aarch64.whl</span><br></pre></td></tr></table></figure>
<p>有可能遇到问题</p>
<figure class="highlight python"><table><tr><td class="code"><pre><span class="line">OSError: libmpi_cxx.so<span class="number">.20</span>: cannot <span class="built_in">open</span> shared <span class="built_in">object</span> file: No such file <span class="keyword">or</span> directory</span><br></pre></td></tr></table></figure>
<p>参考<a href="https://forums.developer.nvidia.com/t/cannot-install-pytorch/149226">Cannot install pytorch</a>，安装两个依赖。</p>
<figure class="highlight bash"><table><tr><td class="code"><pre><span class="line">sudo apt-get install libopenblas-base libopenmpi-dev </span><br></pre></td></tr></table></figure>
<ol start="2">
<li>安装torchvision</li>
</ol>
<p>需要根据torch版本来选择对应版本的torchvision进行安装，参考<a href="https://www.jianshu.com/p/afc502b5a67d">Jetson Nano 安装torch和torchvision</a>。</p>
<figure class="highlight bash"><table><tr><td class="code"><pre><span class="line">git <span class="built_in">clone</span> --branch v0.7.0 https://github.com/pytorch/vision torchvision</span><br><span class="line"><span class="built_in">cd</span> torchvision</span><br><span class="line"><span class="built_in">export</span> BUILD_VERSION=0.7.0</span><br><span class="line">sudo python setup.py install</span><br></pre></td></tr></table></figure>
<ol start="3">
<li>安装mmcv-full和mmdet</li>
</ol>
<p>这里的教程跟<a href="https://github.com/open-mmlab/mmdetection/blob/master/docs/get_started.md">mmdetection官方教程</a>来走就好了。</p>
<p>安装mmcv-full</p>
<figure class="highlight bash"><table><tr><td class="code"><pre><span class="line">git <span class="built_in">clone</span> https://github.com/open-mmlab/mmcv.git</span><br><span class="line"><span class="built_in">cd</span> mmcv</span><br><span class="line">MMCV_WITH_OPS=1 pip install -e . <span class="comment"># 安装full版本</span></span><br></pre></td></tr></table></figure>
<p>安装mmdetection</p>
<figure class="highlight bash"><table><tr><td class="code"><pre><span class="line">git <span class="built_in">clone</span> https://github.com/open-mmlab/mmdetection.git</span><br><span class="line"><span class="built_in">cd</span> mmdetection</span><br><span class="line">pip install -r requirements/build.txt</span><br><span class="line">pip install -v -e .</span><br></pre></td></tr></table></figure>
<p><strong>注意</strong>：</p>
<ul>
<li>mmcv-full和mmdet的版本号要对应，官方教程已经给出来了对应关系。</li>
<li>安装mmcv-full，而不是mmcv，否则后面可能会有奇怪的错误。</li>
</ul>
<figure class="highlight bash"><table><tr><td class="code"><pre><span class="line">MMCV_WITH_OPS=1 pip install -e . -i https://pypi.tuna.tsinghua.edu.cn/simple</span><br></pre></td></tr></table></figure>
<ul>
<li>在使用pip命令时，后面跟上国内源，这样下载其他依赖包会更快一些。</li>
<li>上面两个包安装完，也要花挺长时间的。</li>
<li>有可能遇到安装matplotlib，scipy，pillow等包的时候，与已经安装的numpy版本冲突。这种情况下不能修改numpy的版本，要修改其他包的版本。</li>
<li>可以使用下面这条命令，查看哪个版本的包时候你的numpy，包和版本号可以更改，参考<a href="https://www.cnblogs.com/DLarTisan/p/11749043.html">conda查看某个安装包的依赖项</a>。</li>
</ul>
<figure class="highlight bash"><table><tr><td class="code"><pre><span class="line">conda search scipy=1.0.0 -info</span><br></pre></td></tr></table></figure>
<ol start="4">
<li>测试</li>
</ol>
<p>按照官网的代码，做了一些修改。</p>
<figure class="highlight python"><table><tr><td class="code"><pre><span class="line"><span class="keyword">from</span> mmdet.apis <span class="keyword">import</span> inference_detector, init_detector, show_result_pyplot</span><br><span class="line"></span><br><span class="line">config_file = <span class="string">'configs/faster_rcnn/faster_rcnn_r50_fpn_1x_coco.py'</span></span><br><span class="line"><span class="comment"># download the checkpoint from model zoo and put it in `checkpoints/`</span></span><br><span class="line"><span class="comment"># url: http://download.openmmlab.com/mmdetection/v2.0/faster_rcnn/faster_rcnn_r50_fpn_1x_coco/faster_rcnn_r50_fpn_1x_coco_20200130-047c8118.pth</span></span><br><span class="line">checkpoint_file = <span class="string">'checkpoints/faster_rcnn_r50_fpn_1x_coco_20200130-047c8118.pth'</span></span><br><span class="line">device = <span class="string">'cuda:0'</span></span><br><span class="line"><span class="comment"># init a detector</span></span><br><span class="line">model = init_detector(config_file, checkpoint_file, device=device)</span><br><span class="line"><span class="comment"># inference the demo image</span></span><br><span class="line">result = inference_detector(model, <span class="string">'demo/demo.jpg'</span>)</span><br><span class="line"><span class="comment"># show the results</span></span><br><span class="line">show_result_pyplot(model, <span class="string">'demo/demo.jpg'</span>, result, score_thr=<span class="number">0.3</span>)</span><br></pre></td></tr></table></figure>
<p><strong>参考</strong><br><a href="https://blog.csdn.net/u011119817/article/details/99679350">【Jetson-Nano】jetson_nano安装环境配置及tensorflow和pytorch安装教程</a><br><a href="https://www.jianshu.com/p/6dbf979533ca">Jetson Nano配置与使用（1）开机</a><br><a href="https://zhuanlan.zhihu.com/p/64868319">在Jetson Nano (TX1/TX2)上使用Anaconda与PyTorch 1.1.0</a><br><a href="https://qengineering.eu/install-opencv-4.5-on-jetson-nano.html">Install OpenCV 4.5.0 on Jetson Nano</a><br><a href="https://pythops.com/post/compile-deeplearning-libraries-for-jetson-nano">compile deeplearning libraries for jetson nano</a><br><a href="https://www.cnblogs.com/fenglongyu/p/8654991.html">linux命令系列 sudo apt-get update和upgrade的区别</a><a href="https://github.com/jetsonhacks/buildOpenCVXavier">buildOpenCVXavier</a><br><a href="https://github.com/mdegans/nano_build_opencv">nano_build_opencv</a><br><a href="https://forums.developer.nvidia.com/t/install-opencv-for-python3-in-jetson-nano/74042">在Jetson Nano中为python3安装OpenCV</a><br><a href="https://www.bojankomazec.com/2019/12/installing-opencv-4-on-nvidia-jetson.html">Installing OpenCV 4 on NVIDIA Jetson Nano</a><br><a href="https://blog.csdn.net/weixin_40784980/article/details/105246509">Jetson Nano 从头配置OpenCV+CUDA+QT完整流程</a><br><a href="https://blog.csdn.net/qq_27971677/article/details/90400118">树莓派安装OpenCV-4.1.0及Contrib</a><br><a href="https://www.cnblogs.com/chenzhen0530/p/12109868.html">Ubuntu 18.04配置OpenCV 4.2.0</a><br><a href="https://blog.csdn.net/hongge_smile/article/details/104372783?utm_medium=distribute.pc_relevant_bbs_down.none-task-blog-baidujs-2.nonecase&depth_1-utm_source=distribute.pc_relevant_bbs_down.none-task-blog-baidujs-2.nonecase">opencv安装opencv_contrib出现无法打开包括文件: “opencv2/xfeatures2d/cuda.hpp”</a><br><a href="https://blog.csdn.net/GungnirsPledge/article/details/108597474#5CMakeFilesexample_gpu_surf_keypoint_matcherdirsurf_keypoint_matchercppo_In_function_main_252">源码编译安装OpenCV4.5+ CUDA11 带python 遇到的错误总结</a><br><a href="https://zhuanlan.zhihu.com/p/280702247">cv2: numpy.core.multiarray failed to import</a><br><a href="https://blog.csdn.net/qq_24282081/article/details/107285072#STEP1%20%E5%AE%89%E8%A3%85pytorch%C2%A0">在Jetson tx2安装 mmdetection环境</a></p>
]]></content>
</entry>
<entry>
<title></title>
<url>/2023/04/09/LeetCode%E7%AE%97%E6%B3%95%E9%A2%98%E8%A7%A3%E2%80%94%E2%80%94%E4%BA%8C%E5%88%86%E6%9F%A5%E6%89%BE2/</url>
<content><![CDATA[<p>@<a href="LeetCode%E7%AE%97%E6%B3%95%E9%A2%98%E8%A7%A3%E2%80%94%E2%80%94%E4%BA%8C%E5%88%86%E6%9F%A5%E6%89%BE2">TOC</a></p>
<p>接上文<a href="https://blog.csdn.net/qq_41214610/article/details/105508002">LeetCode算法题解——二分查找1</a>，本篇总结LeetCode中部分题目的二分查找解法。</p>
<h1 id="第四题（寻找左边界）"><a href="#第四题（寻找左边界）" class="headerlink" title="第四题（寻找左边界）"></a>第四题（寻找左边界）</h1><p><a href="https://leetcode.cn/problems/first-bad-version/">278. 第一个错误的版本</a></p>
<p>题目描述：<br>你是产品经理，目前正在带领一个团队开发新的产品。不幸的是，你的产品的最新版本没有通过质量检测。由于每个版本都是基于之前的版本开发的，所以错误的版本之后的所有版本都是错的。</p>
<p>假设你有 n 个版本 [1, 2, …, n]，你想找出导致之后所有版本出错的第一个错误的版本。</p>
<p>你可以通过调用 bool isBadVersion(version) 接口来判断版本号 version 是否在单元测试中出错。实现一个函数来查找第一个错误的版本。你应该尽量减少对调用 API 的次数。</p>
<p>示例1：</p>
<figure class="highlight text"><table><tr><td class="code"><pre><span class="line">输入：n = 5, bad = 4</span><br><span class="line">输出：4</span><br><span class="line">解释：</span><br><span class="line">调用 isBadVersion(3) -> false </span><br><span class="line">调用 isBadVersion(5) -> true </span><br><span class="line">调用 isBadVersion(4) -> true</span><br><span class="line">所以，4 是第一个错误的版本。</span><br></pre></td></tr></table></figure>
<h3 id="思路"><a href="#思路" class="headerlink" title="思路"></a>思路</h3><p>如果将n个版本的true和false放到一个数组里，应该是[false, false, …, true, true]，前面都是false，后面都是true。这道题可以看作是找到最后一个false，也可以看作是找到第一个true。在这里，我将它当作找到第一个true来做，也就是寻找true的左边界。</p>
<h3 id="代码"><a href="#代码" class="headerlink" title="代码"></a>代码</h3><p>java版本</p>
<figure class="highlight java"><table><tr><td class="code"><pre><span class="line"><span class="keyword">public</span> <span class="keyword">class</span> <span class="title class_">Solution</span> <span class="keyword">extends</span> <span class="title class_">VersionControl</span> {</span><br><span class="line"> <span class="keyword">public</span> <span class="type">int</span> <span class="title function_">firstBadVersion</span><span class="params">(<span class="type">int</span> n)</span> {</span><br><span class="line"> <span class="comment">// 左闭右闭</span></span><br><span class="line"> <span class="type">int</span> <span class="variable">l</span> <span class="operator">=</span> <span class="number">1</span>;</span><br><span class="line"> <span class="type">int</span> <span class="variable">r</span> <span class="operator">=</span> n;</span><br><span class="line"> <span class="comment">// 寻找true的左边界</span></span><br><span class="line"> <span class="keyword">while</span>(l <= r){</span><br><span class="line"> <span class="comment">// 防止溢出</span></span><br><span class="line"> <span class="type">int</span> <span class="variable">m</span> <span class="operator">=</span> l + (r - l) / <span class="number">2</span>;</span><br><span class="line"> <span class="comment">// 能找到，缩小右边界</span></span><br><span class="line"> <span class="keyword">if</span>(isBadVersion(m) == <span class="literal">true</span>){</span><br><span class="line"> r = m - <span class="number">1</span>;</span><br><span class="line"> }</span><br><span class="line"> <span class="comment">// 找不到，缩小左边界</span></span><br><span class="line"> <span class="keyword">else</span> <span class="keyword">if</span>(isBadVersion(m) == <span class="literal">false</span>){</span><br><span class="line"> l = m + <span class="number">1</span>;</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> l;</span><br><span class="line"> }</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
<p>python版本</p>
<figure class="highlight python"><table><tr><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span>:</span><br><span class="line"> <span class="keyword">def</span> <span class="title function_">firstBadVersion</span>(<span class="params">self, n: <span class="built_in">int</span></span>) -> <span class="built_in">int</span>:</span><br><span class="line"> l, r = <span class="number">1</span>, n</span><br><span class="line"> <span class="keyword">while</span> l <= r:</span><br><span class="line"> m = l + (r - l) // <span class="number">2</span></span><br><span class="line"> <span class="keyword">if</span> isBadVersion(m):</span><br><span class="line"> r = m - <span class="number">1</span></span><br><span class="line"> <span class="keyword">else</span>:</span><br><span class="line"> l = m + <span class="number">1</span></span><br><span class="line"> <span class="keyword">return</span> l</span><br></pre></td></tr></table></figure>
<h3 id="总结"><a href="#总结" class="headerlink" title="总结"></a>总结</h3><p>在寻找左边界的时候，和之前中点值找到target一样，找到就收缩右边界。拓展一下，在寻找左/右边界，找到了target，但还没有到达其边界，应该向你要搜索的边界靠近，进一步压缩搜索空间。</p>
<h1 id="第五题（寻找右边界）"><a href="#第五题（寻找右边界）" class="headerlink" title="第五题（寻找右边界）"></a>第五题（寻找右边界）</h1><p><a href="https://leetcode.cn/problems/find-smallest-letter-greater-than-target/">744. 寻找比目标字母大的最小字母</a></p>
<p>题目描述：<br>给你一个字符数组 letters，该数组按非递减顺序排序，以及一个字符 target。letters 里至少有两个不同的字符。</p>
<p>返回 letters 中大于 target 的最小的字符。如果不存在这样的字符，则返回 letters 的第一个字符。</p>
<p>示例1：</p>
<figure class="highlight text"><table><tr><td class="code"><pre><span class="line">输入: letters = ["c", "f", "j"]，target = "a"</span><br><span class="line">输出: "c"</span><br><span class="line">解释：letters 中字典上比 'a' 大的最小字符是 'c'。</span><br></pre></td></tr></table></figure>
<h3 id="思路-1"><a href="#思路-1" class="headerlink" title="思路"></a>思路</h3><p>按照题意，寻找第一个比target大的字母。在前面寻找右边界时我们已经提到，当循环停止时右边界r以右的元素都大于target，则右边界r以左的元素都小于等于target，右边界r即最后一个target。所以，第一个比target大的字母就是r+1处的字母。但是当target比数组中所有数字都要大的时候，r+1会越界，这里需要做特别处理。</p>
<h3 id="代码-1"><a href="#代码-1" class="headerlink" title="代码"></a>代码</h3><p>java版本</p>
<figure class="highlight java"><table><tr><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> {</span><br><span class="line"> <span class="keyword">public</span> <span class="type">char</span> <span class="title function_">nextGreatestLetter</span><span class="params">(<span class="type">char</span>[] letters, <span class="type">char</span> target)</span> {</span><br><span class="line"> <span class="comment">// 左闭右闭</span></span><br><span class="line"> <span class="type">int</span> <span class="variable">l</span> <span class="operator">=</span> <span class="number">0</span>;</span><br><span class="line"> <span class="type">int</span> <span class="variable">r</span> <span class="operator">=</span> letters.length - <span class="number">1</span>;</span><br><span class="line"> <span class="comment">// 寻找右边界</span></span><br><span class="line"> <span class="keyword">while</span>(l <= r){</span><br><span class="line"> <span class="comment">// 防止溢出</span></span><br><span class="line"> <span class="type">int</span> <span class="variable">m</span> <span class="operator">=</span> l + (r - l) / <span class="number">2</span> ;</span><br><span class="line"> <span class="comment">// 收缩左边界</span></span><br><span class="line"> <span class="keyword">if</span>(target >= letters[m]){</span><br><span class="line"> l = m + <span class="number">1</span>;</span><br><span class="line"> }</span><br><span class="line"> <span class="comment">// 收缩右边界</span></span><br><span class="line"> <span class="keyword">else</span>{</span><br><span class="line"> r = m - <span class="number">1</span>;</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> <span class="comment">// 防止越界，当不存在这样的字符，即target最大</span></span><br><span class="line"> <span class="keyword">if</span>(r == letters.length - <span class="number">1</span>){</span><br><span class="line"> <span class="keyword">return</span> letters[<span class="number">0</span>];</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> letters[r + <span class="number">1</span>];</span><br><span class="line"> }</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
<p>python版本</p>
<figure class="highlight python"><table><tr><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span>:</span><br><span class="line"> <span class="keyword">def</span> <span class="title function_">nextGreatestLetter</span>(<span class="params">self, letters: <span class="type">List</span>[<span class="built_in">str</span>], target: <span class="built_in">str</span></span>) -> <span class="built_in">str</span>:</span><br><span class="line"> l, r = <span class="number">0</span>, <span class="built_in">len</span>(letters) - <span class="number">1</span></span><br><span class="line"> <span class="keyword">while</span> l <= r:</span><br><span class="line"> m = l + (r - l) // <span class="number">2</span></span><br><span class="line"> <span class="keyword">if</span> target >= letters[m]:</span><br><span class="line"> l = m + <span class="number">1</span></span><br><span class="line"> <span class="keyword">else</span>:</span><br><span class="line"> r = m - <span class="number">1</span></span><br><span class="line"> <span class="keyword">return</span> letters[(r + <span class="number">1</span>) % <span class="built_in">len</span>(letters)]</span><br></pre></td></tr></table></figure>
<h3 id="总结-1"><a href="#总结-1" class="headerlink" title="总结"></a>总结</h3><p>这道题，需要将大于 target 的最小的字符理解成第一个比target大的字符，并且知道可以通过寻找target的右边界来求解。</p>
<h1 id="第六题（找一个数）"><a href="#第六题（找一个数）" class="headerlink" title="第六题（找一个数）"></a>第六题（找一个数）</h1><p><a href="https://leetcode.cn/problems/valid-perfect-square/">367. 有效的完全平方数</a></p>
<p>题目描述：<br>给你一个正整数 num 。如果 num 是一个完全平方数，则返回 true ，否则返回 false 。</p>
<p>完全平方数 是一个可以写成某个整数的平方的整数。换句话说，它可以写成某个整数和自身的乘积。</p>
<p>不能使用任何内置的库函数，如 sqrt 。</p>
<p>示例1：</p>
<figure class="highlight text"><table><tr><td class="code"><pre><span class="line">输入：num = 16</span><br><span class="line">输出：true</span><br><span class="line">解释：返回 true ，因为 4 * 4 = 16 且 4 是一个整数。</span><br></pre></td></tr></table></figure>
<h3 id="思路-2"><a href="#思路-2" class="headerlink" title="思路"></a>思路</h3><p>首先思考一种暴力算法。以num=16为例，我们先从1开始穷举。1*1=1，1<16，不满足。2*2=4，4<16，不满足。3*3=9，9<16，不满足。4*4=16，16==16，满足！5*5=25，25>16，不满足。显然，比4小的数和比4大的数，都不满足，我们可以通过二分法筛选掉比4小的数和比4大的数。</p>
<p>值得注意的是，如果直接比较x*x和num，可能x*x会超过int的范围。可以转化为比较x和num/x。如果是这样的话，考虑到整数除法结果仍然为整数，当x=num/x时，num可能是x*x，x*x+1，…，x*x+(x-1)。此时只需要再次比较x*x和num，即可。</p>
<h3 id="代码-2"><a href="#代码-2" class="headerlink" title="代码"></a>代码</h3><p>java版本</p>
<figure class="highlight java"><table><tr><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> {</span><br><span class="line"> <span class="keyword">public</span> <span class="type">boolean</span> <span class="title function_">isPerfectSquare</span><span class="params">(<span class="type">int</span> num)</span> {</span><br><span class="line"> <span class="comment">// 左闭右闭</span></span><br><span class="line"> <span class="type">int</span> <span class="variable">l</span> <span class="operator">=</span> <span class="number">1</span>;</span><br><span class="line"> <span class="type">int</span> <span class="variable">r</span> <span class="operator">=</span> num;</span><br><span class="line"> <span class="keyword">while</span>(l <= r){</span><br><span class="line"> <span class="comment">// 防止溢出</span></span><br><span class="line"> <span class="type">int</span> <span class="variable">m</span> <span class="operator">=</span> l + (r - l) / <span class="number">2</span>;</span><br><span class="line"> <span class="type">int</span> <span class="variable">sqrt</span> <span class="operator">=</span> num / m;</span><br><span class="line"> <span class="comment">// 找到target并进行判断</span></span><br><span class="line"> <span class="keyword">if</span>((m == sqrt) && (m * sqrt == num)){</span><br><span class="line"> <span class="keyword">return</span> <span class="literal">true</span>;</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">else</span> <span class="keyword">if</span>(m < sqrt){</span><br><span class="line"> l = m + <span class="number">1</span>;</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">else</span>{</span><br><span class="line"> r = m - <span class="number">1</span>;</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> <span class="literal">false</span>;</span><br><span class="line"> }</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
<p>python版本</p>
<figure class="highlight python"><table><tr><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span>:</span><br><span class="line"> <span class="keyword">def</span> <span class="title function_">isPerfectSquare</span>(<span class="params">self, num: <span class="built_in">int</span></span>) -> <span class="built_in">bool</span>: </span><br><span class="line"> l, r = <span class="number">1</span>, num</span><br><span class="line"> <span class="keyword">while</span> l <= r:</span><br><span class="line"> m = l + (r - l) // <span class="number">2</span></span><br><span class="line"> sqrt = num // m</span><br><span class="line"> <span class="keyword">if</span> sqrt < m:</span><br><span class="line"> r = m - <span class="number">1</span></span><br><span class="line"> <span class="keyword">else</span>:</span><br><span class="line"> l = m + <span class="number">1</span></span><br><span class="line"> </span><br><span class="line"> <span class="keyword">return</span> r * r == num</span><br></pre></td></tr></table></figure>
<h3 id="总结-2"><a href="#总结-2" class="headerlink" title="总结"></a>总结</h3><p>这道题，我们可以先思考一个穷举的办法，然后考虑可以用二分法进行优化。防止溢出是一个需要注意的点。这道题里，m是中点，sqrt是target。和以前不同，这道题的target是会变的。</p>
<h1 id="第七题（找右边界）"><a href="#第七题（找右边界）" class="headerlink" title="第七题（找右边界）"></a>第七题（找右边界）</h1><p><a href="https://leetcode.cn/problems/sqrtx/">69. x 的平方根 </a></p>
<p>题目描述：<br>给你一个非负整数 x ，计算并返回 x 的 算术平方根 。</p>
<p>由于返回类型是整数，结果只保留 整数部分 ，小数部分将被 舍去 。</p>
<p>注意：不允许使用任何内置指数函数和算符，例如 pow(x, 0.5) 或者 x ** 0.5 。</p>
<p>示例1：</p>
<figure class="highlight text"><table><tr><td class="code"><pre><span class="line">输入：x = 8</span><br><span class="line">输出：2</span><br><span class="line">解释：8 的算术平方根是 2.82842..., 由于返回类型是整数，小数部分将被舍去。</span><br></pre></td></tr></table></figure>
<h3 id="思路-3"><a href="#思路-3" class="headerlink" title="思路"></a>思路</h3><p>这题和上一题不同。在上一题中，完全平方数，如16，一定能找到整数4*4=16。但是在这题里，将平方根的小数部分舍去了，那么就不能通过x*x=num来进行判断。同样的，我们先思考一种暴力算法。以num=8为例，我们先从1开始穷举。1*1=1，1<8；2*2=4，4<8；3*3=9，9>8。显然，前面的平方数都比8小，后面的平方数都比8大，我们需要找到最后一个比8小的平方数。问题转化为了找比8小的平方数的右边界。</p>
<p>和上一题一样，如果直接比较x*x和num，可能x*x会超过int的范围，可以转化为比较x和num/x。</p>
<h3 id="代码-3"><a href="#代码-3" class="headerlink" title="代码"></a>代码</h3><p>java版本</p>
<figure class="highlight java"><table><tr><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> {</span><br><span class="line"> <span class="keyword">public</span> <span class="type">int</span> <span class="title function_">mySqrt</span><span class="params">(<span class="type">int</span> x)</span> {</span><br><span class="line"> <span class="keyword">if</span>(x <= <span class="number">1</span>){</span><br><span class="line"> <span class="keyword">return</span> x;</span><br><span class="line"> }</span><br><span class="line"> <span class="comment">// 左闭右闭</span></span><br><span class="line"> <span class="type">int</span> <span class="variable">l</span> <span class="operator">=</span> <span class="number">1</span>;</span><br><span class="line"> <span class="type">int</span> <span class="variable">r</span> <span class="operator">=</span> x;</span><br><span class="line"> <span class="keyword">while</span>(l <= r){</span><br><span class="line"> <span class="comment">// 防止溢出</span></span><br><span class="line"> <span class="type">int</span> <span class="variable">m</span> <span class="operator">=</span> l + (r - l) / <span class="number">2</span>;</span><br><span class="line"> <span class="type">int</span> <span class="variable">sqrt</span> <span class="operator">=</span> x / m;</span><br><span class="line"> <span class="keyword">if</span>(m > sqrt){</span><br><span class="line"> r = m - <span class="number">1</span>;</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">else</span>{</span><br><span class="line"> l = m + <span class="number">1</span>;</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> r;</span><br><span class="line"> }</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
<p>python版本</p>
<figure class="highlight python"><table><tr><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span>:</span><br><span class="line"> <span class="keyword">def</span> <span class="title function_">mySqrt</span>(<span class="params">self, x: <span class="built_in">int</span></span>) -> <span class="built_in">int</span>:</span><br><span class="line"> <span class="keyword">if</span> x == <span class="number">0</span>:</span><br><span class="line"> <span class="keyword">return</span> x</span><br><span class="line"></span><br><span class="line"> l, r = <span class="number">1</span>, x</span><br><span class="line"> <span class="keyword">while</span> l <= r:</span><br><span class="line"> m = l + (r - l) // <span class="number">2</span></span><br><span class="line"> sqrt = x // m</span><br><span class="line"> <span class="keyword">if</span> sqrt < m:</span><br><span class="line"> r = m - <span class="number">1</span></span><br><span class="line"> <span class="keyword">else</span>:</span><br><span class="line"> l = m + <span class="number">1</span></span><br><span class="line"> <span class="keyword">return</span> r</span><br></pre></td></tr></table></figure>
<h3 id="总结-3"><a href="#总结-3" class="headerlink" title="总结"></a>总结</h3><p>同样的，这道题我们先思考一个穷举的办法，然后考虑可以用二分法进行优化。防止溢出依旧是一个需要注意的点。这道题里，m是中点，sqrt是target，会发生改变。</p>
<p>下一篇博客<a href="https://blog.csdn.net/qq_41214610/article/details/128875429">LeetCode算法题解——二分查找3</a>中，我将分享LeetCode中几道比较难想到使用二分查找解法的题目。</p>
<p>参考:</p>
<p><a href="https://leetcode-cn.com/problems/find-first-and-last-position-of-element-in-sorted-array/solution/er-fen-cha-zhao-suan-fa-xi-jie-xiang-jie-by-labula/">二分查找算法细节详解，顺便写了首诗</a></p>
]]></content>
</entry>
<entry>
<title></title>
<url>/2023/04/09/LeetCode%E7%AE%97%E6%B3%95%E9%A2%98%E8%A7%A3%E2%80%94%E2%80%94%E4%BA%8C%E5%88%86%E6%9F%A5%E6%89%BE3/</url>
<content><![CDATA[<p>@<a href="LeetCode%E7%AE%97%E6%B3%95%E9%A2%98%E8%A7%A3%E2%80%94%E2%80%94%E4%BA%8C%E5%88%86%E6%9F%A5%E6%89%BE3">TOC</a><br>接上文<a href="https://blog.csdn.net/qq_41214610/article/details/105508002">LeetCode算法题解——二分查找2</a>，本篇分享LeetCode中几道比较难想到使用二分查找解法的题目。</p>
<h1 id="第八题"><a href="#第八题" class="headerlink" title="第八题"></a>第八题</h1><p><a href="https://leetcode.cn/problems/single-element-in-a-sorted-array/">540. 有序数组中的单一元素</a></p>
<p>题目描述：<br>给你一个仅由整数组成的有序数组，其中每个元素都会出现两次，唯有一个数只会出现一次。</p>
<p>请你找出并返回只出现一次的那个数。</p>
<p>你设计的解决方案必须满足 O(log n) 时间复杂度和 O(1) 空间复杂度。</p>
<p>示例1：</p>
<figure class="highlight text"><table><tr><td class="code"><pre><span class="line">输入: nums = [1,1,2,3,3,4,4,8,8]</span><br><span class="line">输出: 2</span><br></pre></td></tr></table></figure>
<h3 id="思路"><a href="#思路" class="headerlink" title="思路"></a>思路</h3><p>我们先思考一下，没有单一元素的nums，如[1,1,3,3,4,4,8,8]具有什么样的规律。不难发现，对于每一个偶数下标i=2k的nums[i]，它的值都等于后一位nums[i+1]的值。但是在插入2之后，这种规律只会在2之前存在。在2之后，所有奇数下标i=2k+1的nums[i]，都等于nums[i+1]。</p>
<p>所以，我们只需要找到中点m附近的偶数下标，通过判断nums[i]=nums[i+1]这种规律是否存在，就可以判断中点m是在插入的单一元素之前还是之后。如果这种规律存在，那么说明中点m在单一元素之前，需要收缩左边界；如果这种规律不存在，那么说明中点m在单一元素之后，需要收缩右边界。</p>
<p>收缩左边界时，因为已经知道nums[m] = nums[m+1]，所以l=m+2。收缩右边界时，nums[m] != nums[m+1]，则必定有nums[m+1]= nums[m+2]，所以r=m。</p>
<p>最后，搜索区间不断缩小，不断将成对的元素排除在区间之外。当区间长度为1时，区间内的元素就是单一元素。</p>
<h3 id="代码"><a href="#代码" class="headerlink" title="代码"></a>代码</h3><p>java版本</p>
<figure class="highlight java"><table><tr><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> {</span><br><span class="line"> <span class="keyword">public</span> <span class="type">int</span> <span class="title function_">singleNonDuplicate</span><span class="params">(<span class="type">int</span>[] nums)</span> {</span><br><span class="line"> <span class="type">int</span> <span class="variable">l</span> <span class="operator">=</span> <span class="number">0</span>, h = nums.length - <span class="number">1</span>;</span><br><span class="line"> <span class="keyword">while</span> (l < h) {</span><br><span class="line"> <span class="comment">// 防止溢出</span></span><br><span class="line"> <span class="type">int</span> <span class="variable">m</span> <span class="operator">=</span> l + (h - l) / <span class="number">2</span>;</span><br><span class="line"> <span class="keyword">if</span> (m % <span class="number">2</span> == <span class="number">1</span>) {</span><br><span class="line"> m--; <span class="comment">// 保证 l/h/m 都在偶数位，使得查找区间大小一直都是奇数</span></span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">if</span> (nums[m] == nums[m + <span class="number">1</span>]) {</span><br><span class="line"> l = m + <span class="number">2</span>;</span><br><span class="line"> } <span class="keyword">else</span> {</span><br><span class="line"> h = m;</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> nums[l];</span><br><span class="line"> }</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
<p>python版本</p>
<figure class="highlight python"><table><tr><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span>:</span><br><span class="line"> <span class="keyword">def</span> <span class="title function_">singleNonDuplicate</span>(<span class="params">self, nums: <span class="type">List</span>[<span class="built_in">int</span>]</span>) -> <span class="built_in">int</span>:</span><br><span class="line"> l, r = <span class="number">0</span>, <span class="built_in">len</span>(nums) - <span class="number">1</span></span><br><span class="line"> <span class="keyword">while</span> l < r:</span><br><span class="line"> m = l + (r - l) // <span class="number">2</span></span><br><span class="line"> <span class="keyword">if</span> m % <span class="number">2</span> == <span class="number">1</span>:</span><br><span class="line"> m -= <span class="number">1</span></span><br><span class="line"> </span><br><span class="line"> <span class="keyword">if</span> nums[m] == nums[m + <span class="number">1</span>]:</span><br><span class="line"> l = m + <span class="number">2</span></span><br><span class="line"> <span class="keyword">else</span>:</span><br><span class="line"> r = m</span><br><span class="line"> <span class="keyword">return</span> nums[l]</span><br></pre></td></tr></table></figure>
<h3 id="总结"><a href="#总结" class="headerlink" title="总结"></a>总结</h3><p>之前的二分查找是查找一个target，这道题的二分查找是查找一种关系。某个点之前，这种关系满足，在某个点之后，这种关系不满足。我们需要通过判断关系的满足与否，缩小搜索区间，找出这个点。</p>
<h1 id="第九题"><a href="#第九题" class="headerlink" title="第九题"></a>第九题</h1><p><a href="https://leetcode.cn/problems/find-minimum-in-rotated-sorted-array/">153. 寻找旋转排序数组中的最小值</a></p>
<p>题目描述：<br>已知一个长度为 n 的数组，预先按照升序排列，经由 1 到 n 次 旋转 后，得到输入数组。例如，原数组 nums = [0,1,2,4,5,6,7] 在变化后可能得到：<br>若旋转 4 次，则可以得到 [4,5,6,7,0,1,2]<br>若旋转 7 次，则可以得到 [0,1,2,4,5,6,7]<br>注意，数组 [a[0], a[1], a[2], …, a[n-1]] 旋转一次 的结果为数组 [a[n-1], a[0], a[1], a[2], …, a[n-2]] 。</p>
<p>给你一个元素值 互不相同 的数组 nums ，它原来是一个升序排列的数组，并按上述情形进行了多次旋转。请你找出并返回数组中的 最小元素 。</p>
<p>你必须设计一个时间复杂度为 O(log n) 的算法解决此问题。</p>
<p>示例1:</p>
<figure class="highlight text"><table><tr><td class="code"><pre><span class="line">输入：nums = [3,4,5,1,2]</span><br><span class="line">输出：1</span><br><span class="line">解释：原数组为 [1,2,3,4,5] ，旋转 3 次得到输入数组。</span><br></pre></td></tr></table></figure>
<h3 id="思路-1"><a href="#思路-1" class="headerlink" title="思路"></a>思路</h3><p>根据单调性，对于升序nums，其任意一点m，其右侧任意一点都大于等于左侧任意一点，即nums[m+] >= nums[m-]。如果发生旋转，则在旋转点k，其右侧任意一点都小于等于左侧任意一点，即nums[k+] <= nums[k-]。通过上一题，我们可以知道，二分查找可以通过缩小搜索区间，找出这种使关系发生变化的点。</p>
<p>旋转点左侧为第一分枝，右侧为第二分枝。我们的目标就是，通过缩小收缩区间，找到第二分枝的左端点，即为旋转点。如果中点m在第二分枝上，根据单调性，有r > m && nums[r] >= nums[m]，那么收缩r=m，使得r保持在第二分枝上。如果r=m-1的话，无法保证还在第二分枝，因为我们只验证过r和m的单调性，没有验证r和m-1的单调性。如果中点m在第一分枝上，则恰好r在旋转点（即第二分枝左端点）右侧，m在旋转点左侧，有nums[r] < nums[m]，那么抛弃中点m及其左边的点，l=m+1。</p>
<h3 id="代码-1"><a href="#代码-1" class="headerlink" title="代码"></a>代码</h3><p>java版本</p>
<figure class="highlight java"><table><tr><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> {</span><br><span class="line"> <span class="keyword">public</span> <span class="type">int</span> <span class="title function_">findMin</span><span class="params">(<span class="type">int</span>[] nums)</span> {</span><br><span class="line"> <span class="type">int</span> <span class="variable">l</span> <span class="operator">=</span> <span class="number">0</span>, h = nums.length - <span class="number">1</span>;</span><br><span class="line"> <span class="keyword">while</span> (l < h) {</span><br><span class="line"> <span class="type">int</span> <span class="variable">m</span> <span class="operator">=</span> l + (h - l) / <span class="number">2</span>;</span><br><span class="line"> <span class="keyword">if</span> (nums[m] <= nums[h]) {</span><br><span class="line"> h = m;</span><br><span class="line"> } <span class="keyword">else</span> {</span><br><span class="line"> l = m + <span class="number">1</span>;</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> nums[l];</span><br><span class="line"> }</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
<p>python版本</p>
<figure class="highlight python"><table><tr><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span>:</span><br><span class="line"> <span class="keyword">def</span> <span class="title function_">findMin</span>(<span class="params">self, nums: <span class="type">List</span>[<span class="built_in">int</span>]</span>) -> <span class="built_in">int</span>:</span><br><span class="line"> l, r = <span class="number">0</span>, <span class="built_in">len</span>(nums) - <span class="number">1</span></span><br><span class="line"> <span class="keyword">while</span> l <= r:</span><br><span class="line"> m = l + (r - l) // <span class="number">2</span></span><br><span class="line"> <span class="keyword">if</span> nums[m] == nums[r]:</span><br><span class="line"> <span class="keyword">break</span></span><br><span class="line"> <span class="keyword">elif</span> nums[r] < nums[m]:</span><br><span class="line"> l = m + <span class="number">1</span></span><br><span class="line"> <span class="keyword">else</span>:</span><br><span class="line"> r = m </span><br><span class="line"> <span class="keyword">return</span> nums[l]</span><br></pre></td></tr></table></figure>
<h3 id="总结-1"><a href="#总结-1" class="headerlink" title="总结"></a>总结</h3><p>上面两道题使用二分查找解题的关键都在于，能够将要找的target与某种关系联系起来。在target的左侧，满足某种关系；在target的右侧，不满足这种关系。</p>
<h1 id="第十题"><a href="#第十题" class="headerlink" title="第十题"></a>第十题</h1><p><a href="https://leetcode.cn/problems/find-minimum-in-rotated-sorted-array-ii/">154. 寻找旋转排序数组中的最小值 II</a></p>
<h3 id="代码-2"><a href="#代码-2" class="headerlink" title="代码"></a>代码</h3><p>python版本</p>
<figure class="highlight python"><table><tr><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span>:</span><br><span class="line"> <span class="keyword">def</span> <span class="title function_">findMin</span>(<span class="params">self, nums: <span class="type">List</span>[<span class="built_in">int</span>]</span>) -> <span class="built_in">int</span>:</span><br><span class="line"> l, r = <span class="number">0</span>, <span class="built_in">len</span>(nums) - <span class="number">1</span></span><br><span class="line"> <span class="keyword">while</span> l <= r:</span><br><span class="line"> m = l + (r - l) // <span class="number">2</span></span><br><span class="line"> <span class="keyword">if</span> nums[r] == nums[m]:</span><br><span class="line"> r -= <span class="number">1</span></span><br><span class="line"> <span class="keyword">elif</span> nums[r] > nums[m]:</span><br><span class="line"> r = m</span><br><span class="line"> <span class="keyword">else</span>:</span><br><span class="line"> l = m + <span class="number">1</span></span><br><span class="line"> <span class="keyword">return</span> nums[l]</span><br></pre></td></tr></table></figure>
<h1 id="第十一题"><a href="#第十一题" class="headerlink" title="第十一题"></a>第十一题</h1><p><a href="https://leetcode.cn/problems/search-a-2d-matrix-ii/">240. 搜索二维矩阵 II</a></p>
<p>题目描述：<br>编写一个高效的算法来搜索 m x n 矩阵 matrix 中的一个目标值 target 。该矩阵具有以下特性：</p>
<p>每行的元素从左到右升序排列。<br>每列的元素从上到下升序排列。</p>
<p>示例1：<br><img src="https://img-blog.csdnimg.cn/48aadf584cf843678ce58565911ae3b2.png" alt="在这里插入图片描述"></p>
<figure class="highlight text"><table><tr><td class="code"><pre><span class="line">输入：matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 5</span><br><span class="line">输出：true</span><br></pre></td></tr></table></figure>
<h3 id="思路-2"><a href="#思路-2" class="headerlink" title="思路"></a>思路</h3><p>在上面两道题中我已经谈到了，可以使用二分查找将target与某种关系联系起来，在target的左侧满足某种关系，而在target的右侧不满足这种关系。在这道题中，这种关系是结构上的。对于任意一个元素，其所在行自左向右递增，所在列自上而下递增。特别的，对于每一行最右的元素m，其左的所在行元素都比它小，其下的所在列元素都比它大。因此，如果给定一个target和m进行比较，target小于m则应该向其左的所在行元素继续比较；target大于m则应该向其下的所在列元素继续比较。如果m的位置超出矩阵的合法范围，则说明找不到target。</p>
<h3 id="代码-3"><a href="#代码-3" class="headerlink" title="代码"></a>代码</h3><p>java版本</p>
<figure class="highlight java"><table><tr><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> {</span><br><span class="line"> <span class="keyword">public</span> <span class="type">boolean</span> <span class="title function_">searchMatrix</span><span class="params">(<span class="type">int</span>[][] matrix, <span class="type">int</span> target)</span> {</span><br><span class="line"> <span class="type">int</span> <span class="variable">m</span> <span class="operator">=</span> matrix.length, n = matrix[<span class="number">0</span>].length;</span><br><span class="line"> <span class="type">int</span> <span class="variable">i</span> <span class="operator">=</span> <span class="number">0</span>, j = n - <span class="number">1</span>;</span><br><span class="line"> <span class="keyword">while</span>(i < m && j >= <span class="number">0</span>){</span><br><span class="line"> <span class="keyword">if</span>(target == matrix[i][j]){</span><br><span class="line"> <span class="keyword">return</span> <span class="literal">true</span>;</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">else</span> <span class="keyword">if</span>(target < matrix[i][j]){</span><br><span class="line"> j--;</span><br><span class="line"> <span class="keyword">continue</span>;</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">else</span>{</span><br><span class="line"> i++;</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> <span class="literal">false</span>; </span><br><span class="line"> }</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
<p>python版本</p>
<figure class="highlight python"><table><tr><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span>:</span><br><span class="line"> <span class="keyword">def</span> <span class="title function_">searchMatrix</span>(<span class="params">self, matrix: <span class="type">List</span>[<span class="type">List</span>[<span class="built_in">int</span>]], target: <span class="built_in">int</span></span>) -> <span class="built_in">bool</span>:</span><br><span class="line"> m = <span class="built_in">len</span>(matrix)</span><br><span class="line"> n = <span class="built_in">len</span>(matrix[<span class="number">0</span>])</span><br><span class="line"> x, y = <span class="number">0</span>, n - <span class="number">1</span></span><br><span class="line"> <span class="keyword">while</span> x < m <span class="keyword">and</span> y >= <span class="number">0</span>:</span><br><span class="line"> <span class="keyword">if</span> matrix[x][y] == target:</span><br><span class="line"> <span class="keyword">return</span> <span class="literal">True</span></span><br><span class="line"> <span class="keyword">elif</span> matrix[x][y] < target:</span><br><span class="line"> x += <span class="number">1</span></span><br><span class="line"> <span class="keyword">else</span>:</span><br><span class="line"> y -= <span class="number">1</span></span><br><span class="line"> </span><br><span class="line"> <span class="keyword">return</span> <span class="literal">False</span></span><br></pre></td></tr></table></figure>
<h3 id="总结-2"><a href="#总结-2" class="headerlink" title="总结"></a>总结</h3><p>这一题和前面两题一样，都需要寻找一种关系作为突破口。在某个点之前，这种关系存在；在某个点之后，这种关系不存在。</p>
<h1 id="第十二题"><a href="#第十二题" class="headerlink" title="第十二题"></a>第十二题</h1><p><a href="https://leetcode.cn/problems/kth-smallest-element-in-a-sorted-matrix/">378. 有序矩阵中第 K 小的元素</a></p>
<p>题目描述：<br>给你一个 n x n 矩阵 matrix ，其中每行和每列元素均按升序排序，找到矩阵中第 k 小的元素。<br>请注意，它是 排序后 的第 k 小元素，而不是第 k 个 不同 的元素。</p>
<p>你必须找到一个内存复杂度优于 O(n2) 的解决方案。</p>
<p>示例1：</p>
<figure class="highlight text"><table><tr><td class="code"><pre><span class="line">输入：matrix = [[1,5,9],[10,11,13],[12,13,15]], k = 8</span><br><span class="line">输出：13</span><br><span class="line">解释：矩阵中的元素为 [1,5,9,10,11,12,13,13,15]，第 8 小元素是 13</span><br></pre></td></tr></table></figure>
<h3 id="思路-3"><a href="#思路-3" class="headerlink" title="思路"></a>思路</h3><p>我们先分析这道题<strong>为什么可以用二分查找来解决</strong>。正如前面三题所示，能够用二分查找解决的问题都存在一种关系，在target之前满足这种关系，在target之后不满足这种关系。让我们来看示例1，在数组[1,5,9,10,11,12,13,13,15]中，如果一个元素是target，那么数组中比它小的元素有8个；如果一个元素小于target，那么数组中比它小的元素小于8个；如果一个元素大于target，那么数组中比它小的元素大于8个。这种关系的存在，使得我们可以用二分查找来解决这道题。在这里，给出一个二分查找的思路作为参考，<a href="https://leetcode.cn/problems/kth-smallest-element-in-a-sorted-matrix/solution/er-fen-chao-ji-jian-dan-by-jacksu1024/">二分，超级简单</a>。</p>
<p>关于这个思路，很多人会纠结的一点是，<strong>如何验证最后求得的l在数组中</strong>？其实就是在问，如果一个元素是target，那么数组中比它小的元素有8个；反过来，如果数组中比一个数小的元素有8个，那么这个元素到底是不是target？不难发现，“一个元素是target，那么数组中比它小的元素有8个”是个充分不必要条件。不妨把数组[1,5,9,10,11,12,13,13,15]最后的15改成30来讨论这个问题。对于数组[1,5,9,10,11,12,13,13,30]，我们还是寻找第8小的元素，正确答案应该是13。但是，在13和30中间，其实还有很多个数满足“数组中比一个数小的元素有8个”这个条件。准确来说，14，15，…，29都满足这个条件。我称这些数为“影子target”。很多人就会纠结，按照<a href="https://leetcode.cn/problems/kth-smallest-element-in-a-sorted-matrix/solution/er-fen-chao-ji-jian-dan-by-jacksu1024/">二分，超级简单</a>的思路，那么如何排除掉这些“影子target”？这里我们就得先想明白，为什么会存在“影子target”。简单来说，“影子target”就是target和target后一位之间的间隙。如果我们把最后一位改成20，那么正确答案还是13，“影子target”是14，15，…，19。如果我们把数组改成[1,5,9,10,11,12,13,16,20]，那么正确答案是16，“影子target”是17，18，19。那么，我们会发现，<strong>16是满足“数组中比一个数小的元素有8个”这个条件的最小的数</strong>。</p>
<p>让我们<strong>从头思考一下这个问题</strong>。<strong>为什么我们可以用二分查找来解决这道问题</strong>？因为在数组中，如果一个元素是target，那么数组中比它小的元素个数cnt等于8；如果一个元素小于target，那么数组中比它小的元素个数cnt小于8；如果一个元素大于target，那么数组中比它小的元素cnt大于8。所以，我们可以用二分查找解决这道题。cnt小于8，则l=m+1；cnt大于8，则r=m-1。<strong>但是，cnt等于8时，我们就找到target了吗</strong>？上面已经分析过了，如果cnt等于8，那么这个元素不一定是target，还有可能是“影子target”。因为在target和target后一位之间，还有间隙。那么我们如何找到正确的target呢？我们已经发现，真实target就是满足cnt等于8的最小值，所以<strong>我们就是在找满足cnt等于8这个条件的左边界</strong>。既然是找左边界，在上一篇博客中我已经讨论过了，其实就是在找到满足cnt等于8这个条件的元素时，不要直接返回结果，而是继续收缩右边界。</p>
<h3 id="代码-4"><a href="#代码-4" class="headerlink" title="代码"></a>代码</h3><figure class="highlight java"><table><tr><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> {</span><br><span class="line"> <span class="keyword">public</span> <span class="type">int</span> <span class="title function_">kthSmallest</span><span class="params">(<span class="type">int</span>[][] matrix, <span class="type">int</span> k)</span> {</span><br><span class="line"> <span class="type">int</span> <span class="variable">n</span> <span class="operator">=</span> matrix.length, l = matrix[<span class="number">0</span>][<span class="number">0</span>], r = matrix[n - <span class="number">1</span>][n - <span class="number">1</span>], cnt = <span class="number">0</span>;</span><br><span class="line"> <span class="keyword">while</span>(l <= r){</span><br><span class="line"> <span class="type">int</span> <span class="variable">m</span> <span class="operator">=</span> l + (r - l) / <span class="number">2</span>;</span><br><span class="line"> cnt = <span class="number">0</span>;</span><br><span class="line"> <span class="keyword">for</span>(<span class="type">int</span> <span class="variable">i</span> <span class="operator">=</span> <span class="number">0</span>; i < n; i++){</span><br><span class="line"> <span class="keyword">for</span>(<span class="type">int</span> <span class="variable">j</span> <span class="operator">=</span> <span class="number">0</span>; j < n && matrix[i][j] <= m; j++){</span><br><span class="line"> cnt++;</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">if</span>(cnt < k){</span><br><span class="line"> l = m + <span class="number">1</span>;</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">else</span>{</span><br><span class="line"> r = m - <span class="number">1</span>;</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> l;</span><br><span class="line"> }</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
<p>python版本</p>
<figure class="highlight python"><table><tr><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span>:</span><br><span class="line"> <span class="keyword">def</span> <span class="title function_">kthSmallest</span>(<span class="params">self, matrix: <span class="type">List</span>[<span class="type">List</span>[<span class="built_in">int</span>]], k: <span class="built_in">int</span></span>) -> <span class="built_in">int</span>:</span><br><span class="line"> n = <span class="built_in">len</span>(matrix)</span><br><span class="line"> l = matrix[<span class="number">0</span>][<span class="number">0</span>]</span><br><span class="line"> r = matrix[n - <span class="number">1</span>][n - <span class="number">1</span>]</span><br><span class="line"> <span class="keyword">while</span> l <= r:</span><br><span class="line"> m = l + (r - l) // <span class="number">2</span></span><br><span class="line"> count = <span class="number">0</span></span><br><span class="line"> <span class="keyword">for</span> row <span class="keyword">in</span> matrix:</span><br><span class="line"> <span class="keyword">for</span> col <span class="keyword">in</span> row:</span><br><span class="line"> <span class="keyword">if</span> col <= m:</span><br><span class="line"> count += <span class="number">1</span></span><br><span class="line"> </span><br><span class="line"> <span class="keyword">if</span> k > count:</span><br><span class="line"> l = m + <span class="number">1</span></span><br><span class="line"> <span class="keyword">else</span>:</span><br><span class="line"> r = m - <span class="number">1</span></span><br><span class="line"></span><br><span class="line"> <span class="keyword">return</span> l</span><br></pre></td></tr></table></figure>
<h3 id="总结-3"><a href="#总结-3" class="headerlink" title="总结"></a>总结</h3><p>这道题不难想到用二分查找来解决。一般人或许可以想到用官方题解差不多的二分法。但是，上面介绍的二分法，需要对为什么可以用二分法，用二分法是查找target还是左/右边界有深刻的理解。</p>
<p>参考：<br><a href="https://github.com/CyC2018/CS-Notes/blob/master/notes/Leetcode%20%E9%A2%98%E8%A7%A3%20-%20%E4%BA%8C%E5%88%86%E6%9F%A5%E6%89%BE.md">Leetcode 题解 - 二分查找</a><br><a href="https://leetcode.cn/problems/kth-smallest-element-in-a-sorted-matrix/solution/er-fen-chao-ji-jian-dan-by-jacksu1024/">二分，超级简单</a></p>
]]></content>
</entry>
<entry>
<title></title>
<url>/2023/04/09/LeetCode%E7%AE%97%E6%B3%95%E9%A2%98%E8%A7%A3%E2%80%94%E2%80%94%E5%8F%8C%E6%8C%87%E9%92%881/</url>
<content><![CDATA[<p>@<a href="LeetCode%E7%AE%97%E6%B3%95%E9%A2%98%E8%A7%A3%E2%80%94%E2%80%94%E5%8F%8C%E6%8C%87%E9%92%881">TOC</a></p>
<p>这里介绍双指针在数组中的第一类题型：一个数组当两个用。</p>
<h1 id="第一题"><a href="#第一题" class="headerlink" title="第一题"></a>第一题</h1><p><a href="https://leetcode.cn/problems/merge-sorted-array/">88. 合并两个有序数组</a></p>
<p>题目描述：<br>给你两个按 非递减顺序 排列的整数数组 nums1 和 nums2，另有两个整数 m 和 n ，分别表示 nums1 和 nums2 中的元素数目。</p>
<p>请你 合并 nums2 到 nums1 中，使合并后的数组同样按 非递减顺序 排列。</p>
<p>注意：最终，合并后数组不应由函数返回，而是存储在数组 nums1 中。为了应对这种情况，nums1 的初始长度为 m + n，其中前 m 个元素表示应合并的元素，后 n 个元素为 0 ，应忽略。nums2 的长度为 n 。</p>
<p>示例1：</p>
<figure class="highlight text"><table><tr><td class="code"><pre><span class="line">输入：nums1 = [1,2,3,0,0,0], m = 3, nums2 = [2,5,6], n = 3</span><br><span class="line">输出：[1,2,2,3,5,6]</span><br><span class="line">解释：需要合并 [1,2,3] 和 [2,5,6] 。</span><br><span class="line">合并结果是 [1,2,2,3,5,6] ，其中斜体加粗标注的为 nums1 中的元素。</span><br></pre></td></tr></table></figure>
<h2 id="思路"><a href="#思路" class="headerlink" title="思路"></a>思路</h2><p>一个简单的想法，借助新建数组nums3来完成nums1和nums2的合并。如果nums1[0]小于nums2[0]，则将nums1[0]放入nums3[0]，继续比较nums1[1]和nums2[0]。同样的，如果nums2[0]小于nums1[0]，则将nums2[0]放入nums3[0]，继续比较nums2[1]和nums1[0]。直到将nums1或nums2遍历完，剩下未遍历的部分可以直接放入nums3。</p>
<p>但是，题目要返回的是nums1而不是nums3。那么我们有两个办法来对上述的方法进行优化。第一个办法，最后遍历nums3，用nums3的元素替换nums1的元素。第二个办法，直接在nums1上进行上述的操作。如果要在nums1上进行上述的操作，那么就需要从大到小比较，先把最大的元素放到nums1最右侧，以免覆盖了未参与比较的nums1元素。</p>
<p><strong>直接把nums1数组当nums3数组来用，就是双指针的精髓所在</strong>。</p>
<h2 id="代码"><a href="#代码" class="headerlink" title="代码"></a>代码</h2><p>java版本</p>
<figure class="highlight java"><table><tr><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> {</span><br><span class="line"> <span class="keyword">public</span> <span class="keyword">void</span> <span class="title function_">merge</span><span class="params">(<span class="type">int</span>[] nums1, <span class="type">int</span> m, <span class="type">int</span>[] nums2, <span class="type">int</span> n)</span> {</span><br><span class="line"> <span class="type">int</span> <span class="variable">idx1</span> <span class="operator">=</span> m-<span class="number">1</span>, idx2 = n - <span class="number">1</span>, idx = m + n - <span class="number">1</span>;</span><br><span class="line"> <span class="keyword">while</span>(idx1>=<span class="number">0</span> || idx2 >=<span class="number">0</span>){</span><br><span class="line"> <span class="keyword">if</span>(idx1 < <span class="number">0</span>){</span><br><span class="line"> nums1[idx--] = nums2[idx2--];</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">else</span> <span class="keyword">if</span>(idx2 < <span class="number">0</span>){</span><br><span class="line"> nums1[idx--] = nums1[idx1--];</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">else</span> <span class="keyword">if</span>(nums1[idx1] > nums2[idx2]){</span><br><span class="line"> nums1[idx--] = nums1[idx1--];</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">else</span>{</span><br><span class="line"> nums1[idx--] = nums2[idx2--]; </span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
<p>python版本</p>
<figure class="highlight python"><table><tr><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span>:</span><br><span class="line"> <span class="keyword">def</span> <span class="title function_">merge</span>(<span class="params">self, nums1: <span class="type">List</span>[<span class="built_in">int</span>], m: <span class="built_in">int</span>, nums2: <span class="type">List</span>[<span class="built_in">int</span>], n: <span class="built_in">int</span></span>) -> <span class="literal">None</span>:</span><br><span class="line"> <span class="string">"""</span></span><br><span class="line"><span class="string"> Do not return anything, modify nums1 in-place instead.</span></span><br><span class="line"><span class="string"> """</span></span><br><span class="line"> idx1, idx2, idx3 = m - <span class="number">1</span>, n - <span class="number">1</span>, m + n - <span class="number">1</span></span><br><span class="line"> <span class="keyword">while</span> idx1 >= <span class="number">0</span> <span class="keyword">and</span> idx2 >= <span class="number">0</span>:</span><br><span class="line"> <span class="keyword">if</span> nums1[idx1] > nums2[idx2]:</span><br><span class="line"> nums1[idx3] = nums1[idx1]</span><br><span class="line"> idx3 -= <span class="number">1</span></span><br><span class="line"> idx1 -= <span class="number">1</span></span><br><span class="line"> <span class="keyword">else</span>:</span><br><span class="line"> nums1[idx3] = nums2[idx2]</span><br><span class="line"> idx3 -= <span class="number">1</span></span><br><span class="line"> idx2 -= <span class="number">1</span></span><br><span class="line"> </span><br><span class="line"> <span class="keyword">while</span> idx3 >= <span class="number">0</span> <span class="keyword">and</span> idx2 >= <span class="number">0</span>:</span><br><span class="line"> nums1[idx3] = nums2[idx2]</span><br><span class="line"> idx3 -= <span class="number">1</span></span><br><span class="line"> idx2 -= <span class="number">1</span></span><br></pre></td></tr></table></figure>
<h1 id="第二题"><a href="#第二题" class="headerlink" title="第二题"></a>第二题</h1><p><a href="https://leetcode.cn/problems/remove-element/">27. 移除元素</a></p>
<p>题目描述：<br>给你一个数组 nums 和一个值 val，你需要 原地 移除所有数值等于 val 的元素，并返回移除后数组的新长度。</p>
<p>不要使用额外的数组空间，你必须仅使用 O(1) 额外空间并 原地 修改输入数组。</p>
<p>元素的顺序可以改变。你不需要考虑数组中超出新长度后面的元素。</p>
<p>示例1：</p>
<figure class="highlight text"><table><tr><td class="code"><pre><span class="line">输入：nums = [3,2,2,3], val = 3</span><br><span class="line">输出：2, nums = [2,2]</span><br><span class="line">解释：函数应该返回新的长度 2, 并且 nums 中的前两个元素均为 2。你不需要考虑数组中超出新长度后面的元素。例如，函数返回的新长度为 2 ，而 nums = [2,2,3,3] 或 nums = [2,2,0,0]，也会被视作正确答案。</span><br></pre></td></tr></table></figure>
<h2 id="思路-1"><a href="#思路-1" class="headerlink" title="思路"></a>思路</h2><p>如果不要求原地修改，那很简单。只需要新建数组nums1，从左到右扫描nums的元素，如果nums的元素不等于val，则将其放入nums1；如果等于val，则跳过。</p>
<p>但是，现在要求原地修改，也就是要<strong>一个数组当作两个用</strong>。和上一题一样，有两种办法。第一种办法，可以先把元素放在数组nums1，然后再用nums1的元素去覆盖nums的元素。第二种办法，可以<strong>直接把nums数组当nums1数组</strong>。但是，这样<strong>会不会导致nums原有的元素被覆盖</strong>呢？实际上是不会的，我们思考一下，如果用idx1扫描nums，非val元素通过idx2存放到nums1，那么<strong>idx1总是大于等于idx2</strong>的。也就是说，<strong>idx2使用的区域，idx1早就扫描过了</strong>。</p>
<h2 id="代码-1"><a href="#代码-1" class="headerlink" title="代码"></a>代码</h2><p>java版本</p>
<figure class="highlight java"><table><tr><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> {</span><br><span class="line"> <span class="keyword">public</span> <span class="type">int</span> <span class="title function_">removeElement</span><span class="params">(<span class="type">int</span>[] nums, <span class="type">int</span> val)</span> {</span><br><span class="line"> <span class="type">int</span> <span class="variable">slow</span> <span class="operator">=</span> <span class="number">0</span>;</span><br><span class="line"> <span class="type">int</span> <span class="variable">fast</span> <span class="operator">=</span> <span class="number">0</span>;</span><br><span class="line"> <span class="keyword">for</span>(; fast < nums.length; fast++){</span><br><span class="line"> <span class="keyword">if</span>(nums[fast] != val){</span><br><span class="line"> nums[slow] = nums[fast];</span><br><span class="line"> slow++;</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> slow;</span><br><span class="line"> }</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
<p>python版本</p>
<figure class="highlight python"><table><tr><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span>:</span><br><span class="line"> <span class="keyword">def</span> <span class="title function_">removeElement</span>(<span class="params">self, nums: <span class="type">List</span>[<span class="built_in">int</span>], val: <span class="built_in">int</span></span>) -> <span class="built_in">int</span>:</span><br><span class="line"> n = <span class="built_in">len</span>(nums)</span><br><span class="line"> slow, fast = <span class="number">0</span>, <span class="number">0</span></span><br><span class="line"> <span class="keyword">while</span> fast < n:</span><br><span class="line"> <span class="keyword">if</span> nums[fast] != val:</span><br><span class="line"> nums[slow] = nums[fast]</span><br><span class="line"> slow += <span class="number">1</span></span><br><span class="line"> fast += <span class="number">1</span></span><br><span class="line"> <span class="keyword">return</span> slow</span><br></pre></td></tr></table></figure>
<h1 id="第三题"><a href="#第三题" class="headerlink" title="第三题"></a>第三题</h1><p><a href="https://leetcode.cn/problems/move-zeroes/">283. 移动零</a></p>
<p>题目描述：<br>给定一个数组 nums，编写一个函数将所有 0 移动到数组的末尾，同时保持非零元素的相对顺序。</p>
<p>请注意 ，必须在不复制数组的情况下原地对数组进行操作。</p>
<p>示例1：</p>
<figure class="highlight text"><table><tr><td class="code"><pre><span class="line">输入: nums = [0,1,0,3,12]</span><br><span class="line">输出: [1,3,12,0,0]</span><br></pre></td></tr></table></figure>
<h2 id="思路-2"><a href="#思路-2" class="headerlink" title="思路"></a>思路</h2><p>这题和上一题很相似，都是遇到val（0）就跳过。但是不同的是，这道题不仅要跳过val，还要把val放到后面去。有两种办法，第一种是和第二题一样，遇到val的时候就跳过val，然后再把后面都填充为val；第二种是让val和非val进行交换。</p>
<h2 id="代码-2"><a href="#代码-2" class="headerlink" title="代码"></a>代码</h2><p>java版本</p>
<figure class="highlight java"><table><tr><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> {</span><br><span class="line"> <span class="keyword">public</span> <span class="keyword">void</span> <span class="title function_">moveZeroes</span><span class="params">(<span class="type">int</span>[] nums)</span> {</span><br><span class="line"> <span class="type">int</span> <span class="variable">slow</span> <span class="operator">=</span> <span class="number">0</span>;</span><br><span class="line"> <span class="type">int</span> <span class="variable">fast</span> <span class="operator">=</span> <span class="number">0</span>;</span><br><span class="line"> <span class="keyword">final</span> <span class="type">int</span> <span class="variable">val</span> <span class="operator">=</span> <span class="number">0</span>;</span><br><span class="line"> <span class="keyword">for</span>(; fast < nums.length; fast++){</span><br><span class="line"> <span class="keyword">if</span>(nums[fast] != val){</span><br><span class="line"> <span class="type">int</span> <span class="variable">temp</span> <span class="operator">=</span> nums[slow];</span><br><span class="line"> nums[slow] = nums[fast];</span><br><span class="line"> nums[fast] = temp;</span><br><span class="line"> slow++;</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
<p>python版本</p>
<figure class="highlight python"><table><tr><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span>:</span><br><span class="line"> <span class="keyword">def</span> <span class="title function_">moveZeroes</span>(<span class="params">self, nums: <span class="type">List</span>[<span class="built_in">int</span>]</span>) -> <span class="literal">None</span>:</span><br><span class="line"> <span class="string">"""</span></span><br><span class="line"><span class="string"> Do not return anything, modify nums in-place instead.</span></span><br><span class="line"><span class="string"> """</span></span><br><span class="line"> n = <span class="built_in">len</span>(nums)</span><br><span class="line"> slow, fast = <span class="number">0</span>, <span class="number">0</span></span><br><span class="line"> <span class="keyword">while</span> fast < n:</span><br><span class="line"> <span class="keyword">if</span> nums[fast] != <span class="number">0</span>:</span><br><span class="line"> nums[slow], nums[fast] = nums[fast], nums[slow]</span><br><span class="line"> slow += <span class="number">1</span></span><br><span class="line"> fast += <span class="number">1</span></span><br></pre></td></tr></table></figure>
<h1 id="第四题"><a href="#第四题" class="headerlink" title="第四题"></a>第四题</h1><p><a href="https://leetcode.cn/problems/remove-duplicates-from-sorted-array/">26. 删除有序数组中的重复项</a></p>
<p>题目描述：<br>给你一个 升序排列 的数组 nums ，请你 原地 删除重复出现的元素，使每个元素 只出现一次 ，返回删除后数组的新长度。元素的 相对顺序 应该保持 一致 。</p>
<p>由于在某些语言中不能改变数组的长度，所以必须将结果放在数组nums的第一部分。更规范地说，如果在删除重复项之后有 k 个元素，那么 nums 的前 k 个元素应该保存最终结果。</p>
<p>将最终结果插入 nums 的前 k 个位置后返回 k 。</p>
<p>不要使用额外的空间，你必须在 原地 修改输入数组 并在使用 O(1) 额外空间的条件下完成。</p>
<p>示例1：</p>
<figure class="highlight text"><table><tr><td class="code"><pre><span class="line">输入：nums = [1,1,2]</span><br><span class="line">输出：2, nums = [1,2,_]</span><br><span class="line">解释：函数应该返回新的长度 2 ，并且原数组 nums 的前两个元素被修改为 1, 2 。不需要考虑数组中超出新长度后面的元素。</span><br></pre></td></tr></table></figure>
<h2 id="思路-3"><a href="#思路-3" class="headerlink" title="思路"></a>思路</h2><p>如果不是原地修改，那我们还是考虑新建数组nums1。nums[0]肯定是不重复的，可以放到nums1[0]。因为nums是有序的，从nums[1]开始，如果和前面的元素相同，那么说明是重复的，可以跳过；如果和前面的元素不同，那么说明不重复，放入nums1数组。</p>
<p>如果要原地修改，还是同样的情况，有两种办法。第一种办法就是用nums1去覆盖nums。第二种办法，要<strong>一个数组当作两个用</strong>。我们总是要思考，<strong>会不会覆盖未扫描的元素</strong>？看得出，fast总是比slow跑得快，所以不会覆盖。</p>
<h2 id="代码-3"><a href="#代码-3" class="headerlink" title="代码"></a>代码</h2><p>java版本</p>
<figure class="highlight java"><table><tr><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> {</span><br><span class="line"> <span class="keyword">public</span> <span class="type">int</span> <span class="title function_">removeDuplicates</span><span class="params">(<span class="type">int</span>[] nums)</span> {</span><br><span class="line"> <span class="type">int</span> <span class="variable">slow</span> <span class="operator">=</span> <span class="number">0</span>;</span><br><span class="line"> <span class="type">int</span> <span class="variable">fast</span> <span class="operator">=</span> <span class="number">0</span>;</span><br><span class="line"> <span class="keyword">for</span>(; fast < nums.length; fast++){</span><br><span class="line"> <span class="keyword">if</span>(nums[fast] != nums[slow]){</span><br><span class="line"> slow += <span class="number">1</span>;</span><br><span class="line"> nums[slow] = nums[fast];</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"></span><br><span class="line"> <span class="keyword">return</span> slow + <span class="number">1</span>;</span><br><span class="line"> }</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
<p>python版本</p>
<figure class="highlight python"><table><tr><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span>:</span><br><span class="line"> <span class="keyword">def</span> <span class="title function_">removeDuplicates</span>(<span class="params">self, nums: <span class="type">List</span>[<span class="built_in">int</span>]</span>) -> <span class="built_in">int</span>:</span><br><span class="line"> n = <span class="built_in">len</span>(nums)</span><br><span class="line"> slow, fast = <span class="number">0</span>, <span class="number">0</span></span><br><span class="line"> <span class="keyword">while</span> fast < n:</span><br><span class="line"> <span class="keyword">if</span> nums[fast] != nums[slow]:</span><br><span class="line"> slow += <span class="number">1</span></span><br><span class="line"> nums[slow] = nums[fast]</span><br><span class="line"></span><br><span class="line"> fast += <span class="number">1</span></span><br><span class="line"> </span><br><span class="line"> <span class="keyword">return</span> slow + <span class="number">1</span></span><br></pre></td></tr></table></figure>
<p>下一篇博客<a href="">LeetCode算法题解——双指针2</a>中，我将分享LeetCode中另一种类型的双指针问题。</p>
<p>参考：</p>
<p><a href="https://github.com/CyC2018/CS-Notes/blob/master/notes/Leetcode%20%E9%A2%98%E8%A7%A3%20-%20%E5%8F%8C%E6%8C%87%E9%92%88.md">Leetcode 题解 - 双指针</a></p>
]]></content>
</entry>
<entry>
<title></title>
<url>/2023/04/09/LeetCode%E7%AE%97%E6%B3%95%E9%A2%98%E8%A7%A3%E2%80%94%E2%80%94%E4%BA%8C%E5%88%86%E6%9F%A5%E6%89%BE1/</url>
<content><![CDATA[<p>@<a href="LeetCode%E7%AE%97%E6%B3%95%E9%A2%98%E8%A7%A3%E2%80%94%E2%80%94%E4%BA%8C%E5%88%86%E6%9F%A5%E6%89%BE1">TOC</a><br>常用的二分查找包括：寻找一个数、插入target、寻找左右边界。</p>
<h1 id="第一题（寻找一个数）"><a href="#第一题（寻找一个数）" class="headerlink" title="第一题（寻找一个数）"></a>第一题（寻找一个数）</h1><p><a href="https://leetcode.cn/problems/binary-search/">704. 二分查找</a></p>
<p>题目描述：<br>给定一个 n 个元素有序的（升序）整型数组 nums 和一个目标值 target ，写一个函数搜索 nums 中的 target，如果目标值存在返回下标，否则返回 -1。</p>
<p>示例1：</p>
<figure class="highlight text"><table><tr><td class="code"><pre><span class="line">输入: nums = [-1,0,3,5,9,12], target = 9</span><br><span class="line">输出: 4</span><br><span class="line">解释: 9 出现在 nums 中并且下标为 4</span><br></pre></td></tr></table></figure>
<h3 id="思路"><a href="#思路" class="headerlink" title="思路"></a>思路</h3><p>二分法，如果找到target，直接返回；如果target比中点值小，说明在左区间，则缩小右端点；如果target比中点值大，说明在右区间，则缩小左端点。</p>
<h3 id="代码"><a href="#代码" class="headerlink" title="代码"></a>代码</h3><p>c++版本</p>
<figure class="highlight cpp"><table><tr><td class="code"><pre><span class="line"><span class="function"><span class="type">int</span> <span class="title">binarySearch</span><span class="params">(vector<<span class="type">int</span>> nums, <span class="type">int</span> target)</span> </span></span><br><span class="line"><span class="function"></span>{</span><br><span class="line"> <span class="comment">// 查找区间</span></span><br><span class="line"> <span class="type">int</span> l = <span class="number">0</span>; </span><br><span class="line"> <span class="type">int</span> r = nums.<span class="built_in">size</span>() - <span class="number">1</span>;</span><br><span class="line"> <span class="comment">// 终止条件</span></span><br><span class="line"> <span class="keyword">while</span>(l <= r) </span><br><span class="line"> {</span><br><span class="line"> <span class="comment">// 中点计算</span></span><br><span class="line"> <span class="type">int</span> m = l + (r - l) / <span class="number">2</span>;</span><br><span class="line"> <span class="comment">// 区间缩减</span></span><br><span class="line"> <span class="keyword">if</span>(nums[m] == target)</span><br><span class="line"> <span class="keyword">return</span> m; </span><br><span class="line"> <span class="keyword">else</span> <span class="keyword">if</span> (nums[m] < target)</span><br><span class="line"> l = m + <span class="number">1</span>;</span><br><span class="line"> <span class="keyword">else</span> <span class="keyword">if</span> (nums[m] > target)</span><br><span class="line"> r = m - <span class="number">1</span>;</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> <span class="number">-1</span>;</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
<p>java版本</p>
<figure class="highlight java"><table><tr><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> {</span><br><span class="line"> <span class="keyword">public</span> <span class="type">int</span> <span class="title function_">search</span><span class="params">(<span class="type">int</span>[] nums, <span class="type">int</span> target)</span> {</span><br><span class="line"> <span class="comment">// 左闭右闭</span></span><br><span class="line"> <span class="type">int</span> <span class="variable">l</span> <span class="operator">=</span> <span class="number">0</span>;</span><br><span class="line"> <span class="type">int</span> <span class="variable">r</span> <span class="operator">=</span> nums.length - <span class="number">1</span>;</span><br><span class="line"> <span class="keyword">while</span>(l <= r){</span><br><span class="line"> <span class="comment">// 防止溢出</span></span><br><span class="line"> <span class="type">int</span> <span class="variable">m</span> <span class="operator">=</span> l + (r - l) / <span class="number">2</span>;</span><br><span class="line"> <span class="keyword">if</span>(target == nums[m]){</span><br><span class="line"> <span class="comment">// 如果找到，直接返回</span></span><br><span class="line"> <span class="keyword">return</span> m;</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">else</span> <span class="keyword">if</span>(target > nums[m]){</span><br><span class="line"> <span class="comment">// 右边界缩小</span></span><br><span class="line"> r = m - <span class="number">1</span>;</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">else</span> <span class="keyword">if</span>(target > nums[m]){</span><br><span class="line"> <span class="comment">// 左边界缩小</span></span><br><span class="line"> l = m + <span class="number">1</span>;</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> <span class="comment">// 未找到，返回-1</span></span><br><span class="line"> <span class="keyword">return</span> -<span class="number">1</span>;</span><br><span class="line"> }</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
<p>python版本</p>
<figure class="highlight python"><table><tr><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span>:</span><br><span class="line"> <span class="keyword">def</span> <span class="title function_">search</span>(<span class="params">self, nums: <span class="type">List</span>[<span class="built_in">int</span>], target: <span class="built_in">int</span></span>) -> <span class="built_in">int</span>:</span><br><span class="line"> l, r = <span class="number">0</span>, <span class="built_in">len</span>(nums) - <span class="number">1</span></span><br><span class="line"> <span class="keyword">while</span> l <= r:</span><br><span class="line"> m = l + (r - l) // <span class="number">2</span></span><br><span class="line"> <span class="keyword">if</span> target == nums[m]:</span><br><span class="line"> <span class="keyword">return</span> m</span><br><span class="line"> <span class="keyword">elif</span> target < nums[m]:</span><br><span class="line"> r = m - <span class="number">1</span></span><br><span class="line"> <span class="keyword">else</span>:</span><br><span class="line"> l = m + <span class="number">1</span></span><br><span class="line"> <span class="keyword">return</span> -<span class="number">1</span></span><br></pre></td></tr></table></figure>
<h3 id="总结"><a href="#总结" class="headerlink" title="总结"></a>总结</h3><p>本题有两点需要注意。<br>第一点，搜索区间是“左闭右闭”。之所以选择“左闭右闭”的搜索区间，是为了在缩小左、右端点的时候可以统一操作。此外，因为搜索区间是“左闭右闭”的，所以循环条件是左端点小于等于右端点。<br>第二点，取中点的计算要防溢出。</p>
<h1 id="第二题（插入target）"><a href="#第二题（插入target）" class="headerlink" title="第二题（插入target）"></a>第二题（插入target）</h1><p><a href="https://leetcode.cn/problems/search-insert-position/">35. 搜索插入位置</a></p>
<p>题目描述：<br>给定一个排序数组和一个目标值，在数组中找到目标值，并返回其索引。如果目标值不存在于数组中，返回它将会被按顺序插入的位置。</p>
<p>请必须使用时间复杂度为 O(log n) 的算法。</p>
<p>示例1：</p>
<figure class="highlight text"><table><tr><td class="code"><pre><span class="line">输入: nums = [1,3,5,6], target = 2</span><br><span class="line">输出: 1</span><br></pre></td></tr></table></figure>
<h3 id="思路-1"><a href="#思路-1" class="headerlink" title="思路"></a>思路</h3><p>如果找得到，则参考上题代码。如果找不到，循环停止时，左边界l以左（不包含l）的元素都小于target，右边界r以右（不包含r）的元素都大于target。所以，target应该插入左边界l的位置，这样其左侧元素皆小于target，其右侧元素皆大于target。</p>
<h3 id="代码-1"><a href="#代码-1" class="headerlink" title="代码"></a>代码</h3><p>java版本</p>
<figure class="highlight java"><table><tr><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> {</span><br><span class="line"> <span class="keyword">public</span> <span class="type">int</span> <span class="title function_">searchInsert</span><span class="params">(<span class="type">int</span>[] nums, <span class="type">int</span> target)</span> {</span><br><span class="line"> <span class="comment">// 左闭右闭</span></span><br><span class="line"> <span class="type">int</span> <span class="variable">l</span> <span class="operator">=</span> <span class="number">0</span>;</span><br><span class="line"> <span class="type">int</span> <span class="variable">r</span> <span class="operator">=</span> nums.length - <span class="number">1</span>;</span><br><span class="line"> <span class="keyword">while</span>(l <= r){</span><br><span class="line"> <span class="comment">// 防止溢出</span></span><br><span class="line"> <span class="type">int</span> <span class="variable">m</span> <span class="operator">=</span> l + (r - l)/<span class="number">2</span>;</span><br><span class="line"> <span class="keyword">if</span>(target == nums[m]){</span><br><span class="line"> <span class="comment">// 如果找到，直接返回</span></span><br><span class="line"> <span class="keyword">return</span> m;</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">else</span> <span class="keyword">if</span>(target > nums[m]){</span><br><span class="line"> <span class="comment">// 右边界缩小</span></span><br><span class="line"> r = m - <span class="number">1</span>;</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">else</span> <span class="keyword">if</span>(target > nums[m]){</span><br><span class="line"> <span class="comment">// 左边界缩小</span></span><br><span class="line"> l = m + <span class="number">1</span>;</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> <span class="comment">// 返回左边界</span></span><br><span class="line"> <span class="keyword">return</span> l;</span><br><span class="line"> }</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
<p>python版本</p>
<figure class="highlight python"><table><tr><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span>:</span><br><span class="line"> <span class="keyword">def</span> <span class="title function_">searchInsert</span>(<span class="params">self, nums: <span class="type">List</span>[<span class="built_in">int</span>], target: <span class="built_in">int</span></span>) -> <span class="built_in">int</span>:</span><br><span class="line"> l, r = <span class="number">0</span>, <span class="built_in">len</span>(nums) - <span class="number">1</span></span><br><span class="line"> <span class="keyword">while</span> l <= r:</span><br><span class="line"> m = l + (r - l) // <span class="number">2</span></span><br><span class="line"> <span class="keyword">if</span> nums[m] == target:</span><br><span class="line"> <span class="keyword">return</span> m</span><br><span class="line"> <span class="keyword">elif</span> target < nums[m]:</span><br><span class="line"> r = m - <span class="number">1</span></span><br><span class="line"> <span class="keyword">else</span>:</span><br><span class="line"> l = m + <span class="number">1</span></span><br><span class="line"> <span class="keyword">return</span> l</span><br></pre></td></tr></table></figure>
<h3 id="总结-1"><a href="#总结-1" class="headerlink" title="总结"></a>总结</h3><p>循环停止时，左边界l以左（不包含l）的元素都小于target，右边界r以右（不包含r）的元素都大于target。所以，target应该插入左边界l的位置，这样其左侧元素皆小于target，其右侧元素皆大于target。</p>
<h1 id="第三题（寻找左右边界）"><a href="#第三题（寻找左右边界）" class="headerlink" title="第三题（寻找左右边界）"></a>第三题（寻找左右边界）</h1><p><a href="https://leetcode.cn/problems/find-first-and-last-position-of-element-in-sorted-array/">34. 在排序数组中查找元素的第一个和最后一个位置</a></p>
<p>题目描述：<br>给你一个按照非递减顺序排列的整数数组 nums，和一个目标值 target。请你找出给定目标值在数组中的开始位置和结束位置。</p>
<p>如果数组中不存在目标值 target，返回 [-1, -1]。</p>
<p>你必须设计并实现时间复杂度为 O(log n) 的算法解决此问题。</p>
<p>示例1：</p>
<figure class="highlight text"><table><tr><td class="code"><pre><span class="line">输入：nums = [5,7,7,8,8,10], target = 8</span><br><span class="line">输出：[3,4]</span><br></pre></td></tr></table></figure>
<h3 id="思路-2"><a href="#思路-2" class="headerlink" title="思路"></a>思路</h3><p>先考虑左边界。和前面两题不同，这题找到target不能直接返回，要继续寻找，直到找到第一个target。当循环停止时，左边界l以左的元素都小于target，则左边界l以右的元素都大于等于target，左边界l即第一个target。所以，当找到target的时候，仍然缩小右边界，向左边界靠拢。</p>
<p>同理，搜索右边界时，当找到target不能直接返回，要继续寻找，直到找到最后一个target。当循环停止时，右边界r以右的元素都大于target，则右边界r以左的元素都小于等于target，右边界r即最后一个target。所以，当找到target的时候，仍然缩小左边界，向右边界靠拢。</p>
<p>得到了左右边界，也只是说明如果有target，那第一个和最后一个target就是左右边界。但是，存在没有target的可能，所以还需要对左右边界进行判断。值得注意的是，可能target比所有的元素都大，此时左边界l等于数组长度，需要先判断是否越界，再判断左边界是否为target。同样的，可能target比所有的元素都小，此时右边界l等于-1，需要先判断是否越界，再判断右边界是否为target。</p>
<h3 id="代码-2"><a href="#代码-2" class="headerlink" title="代码"></a>代码</h3><p>java版本</p>
<figure class="highlight java"><table><tr><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> {</span><br><span class="line"> <span class="keyword">public</span> <span class="type">int</span>[] searchRange(<span class="type">int</span>[] nums, <span class="type">int</span> target) {</span><br><span class="line"> <span class="type">int</span> results[] = <span class="keyword">new</span> <span class="title class_">int</span>[]{-<span class="number">1</span>, -<span class="number">1</span>};</span><br><span class="line"> <span class="comment">// 左闭右闭</span></span><br><span class="line"> <span class="type">int</span> <span class="variable">l</span> <span class="operator">=</span> <span class="number">0</span>;</span><br><span class="line"> <span class="type">int</span> <span class="variable">r</span> <span class="operator">=</span> nums.length - <span class="number">1</span>;</span><br><span class="line"> <span class="comment">// 寻找第一个target</span></span><br><span class="line"> <span class="keyword">while</span>(l <= r){</span><br><span class="line"> <span class="comment">// 防止溢出</span></span><br><span class="line"> <span class="type">int</span> <span class="variable">m</span> <span class="operator">=</span> l + (r - l) / <span class="number">2</span>;</span><br><span class="line"> <span class="comment">// 缩小右边界</span></span><br><span class="line"> <span class="keyword">if</span>(target <= nums[m]){</span><br><span class="line"> r = m - <span class="number">1</span>;</span><br><span class="line"> }</span><br><span class="line"> <span class="comment">// 缩小左边界</span></span><br><span class="line"> <span class="keyword">else</span> <span class="keyword">if</span>(target > nums[m]){</span><br><span class="line"> l = m + <span class="number">1</span>;</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> <span class="comment">// 防止越界</span></span><br><span class="line"> <span class="keyword">if</span>((l < nums.length) && (nums[l] == target)){</span><br><span class="line"> results[<span class="number">0</span>] = l;</span><br><span class="line"> }</span><br><span class="line"> <span class="comment">// 左闭右闭</span></span><br><span class="line"> l = <span class="number">0</span>;</span><br><span class="line"> r = nums.length - <span class="number">1</span>;</span><br><span class="line"> <span class="comment">// 寻找最后一个target</span></span><br><span class="line"> <span class="keyword">while</span>(l <= r){</span><br><span class="line"> <span class="comment">// 防止溢出</span></span><br><span class="line"> <span class="type">int</span> <span class="variable">m</span> <span class="operator">=</span> l + (r - l) / <span class="number">2</span>;</span><br><span class="line"> <span class="comment">// 缩小左边界</span></span><br><span class="line"> <span class="keyword">if</span>(target >= nums[m]){</span><br><span class="line"> l = m + <span class="number">1</span>;</span><br><span class="line"> }</span><br><span class="line"> <span class="comment">// 缩小右边界</span></span><br><span class="line"> <span class="keyword">else</span> <span class="keyword">if</span>(target < nums[m]){</span><br><span class="line"> r = m - <span class="number">1</span>;</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> <span class="comment">// 防止越界</span></span><br><span class="line"> <span class="keyword">if</span>((r >= <span class="number">0</span>) && (nums[r] == target)){</span><br><span class="line"> results[<span class="number">1</span>] = r;</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> results;</span><br><span class="line"> }</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
<p>python版本</p>
<figure class="highlight python"><table><tr><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span>:</span><br><span class="line"> <span class="keyword">def</span> <span class="title function_">searchRange</span>(<span class="params">self, nums: <span class="type">List</span>[<span class="built_in">int</span>], target: <span class="built_in">int</span></span>) -> <span class="type">List</span>[<span class="built_in">int</span>]:</span><br><span class="line"> <span class="keyword">if</span> nums == <span class="literal">None</span> <span class="keyword">or</span> <span class="built_in">len</span>(nums) == <span class="number">0</span>:</span><br><span class="line"> <span class="keyword">return</span> [-<span class="number">1</span>, -<span class="number">1</span>]</span><br><span class="line"> </span><br><span class="line"> result = [-<span class="number">1</span>, -<span class="number">1</span>]</span><br><span class="line"></span><br><span class="line"> l, r = <span class="number">0</span>, <span class="built_in">len</span>(nums) - <span class="number">1</span></span><br><span class="line"> <span class="keyword">while</span> l <= r:</span><br><span class="line"> m = l + (r - l) // <span class="number">2</span></span><br><span class="line"> <span class="keyword">if</span> target <= nums[m]:</span><br><span class="line"> r = m - <span class="number">1</span></span><br><span class="line"> <span class="keyword">else</span>:</span><br><span class="line"> l = m + <span class="number">1</span></span><br><span class="line"> </span><br><span class="line"> <span class="keyword">if</span> l < <span class="built_in">len</span>(nums) <span class="keyword">and</span> nums[l] == target:</span><br><span class="line"> result[<span class="number">0</span>] = l</span><br><span class="line"> </span><br><span class="line"> l, r = <span class="number">0</span>, <span class="built_in">len</span>(nums) - <span class="number">1</span></span><br><span class="line"> <span class="keyword">while</span> l <= r:</span><br><span class="line"> m = l + (r - l) // <span class="number">2</span></span><br><span class="line"> <span class="keyword">if</span> target >= nums[m]:</span><br><span class="line"> l = m + <span class="number">1</span></span><br><span class="line"> <span class="keyword">else</span>:</span><br><span class="line"> r = m - <span class="number">1</span></span><br><span class="line"> </span><br><span class="line"> <span class="keyword">if</span> r >= <span class="number">0</span> <span class="keyword">and</span> nums[r] == target:</span><br><span class="line"> result[<span class="number">1</span>] = r</span><br><span class="line"></span><br><span class="line"> <span class="keyword">return</span> result</span><br></pre></td></tr></table></figure>
<h3 id="总结-2"><a href="#总结-2" class="headerlink" title="总结"></a>总结</h3><p>本题有三点需要注意。</p>
<p>第一点，理解为什么左右边界就是第一个和最后一个target。当循环停止时，左边界l以左的元素都小于target，则左边界l以右的元素都大于等于target，左边界l即第一个target。当循环停止时，右边界r以右的元素都大于target，则右边界r以左的元素都小于等于target，右边界r即最后一个target。</p>
<p>第二点，当找到target的时候左右边界如何移动。寻找左边界时，需要右边界向左边界靠拢，所以找到target时仍然缩小右边界。寻找右边界时，需要左边界向右边界靠拢，所以找到target时仍然缩小左边界。</p>
<p>第三点，防止左右边界越界。左右边界是target“应该”出现的第一个和最后一个位置，到底是不是还需要经过验证。验证之前，记得考虑左右边界是否越界。</p>
<p>下一篇博客<a href="https://blog.csdn.net/qq_41214610/article/details/128852619">LeetCode算法题解——二分查找2</a>，我将总结LeetCode中部分题目的二分查找解法。</p>
<p>参考:</p>
<p><a href="https://leetcode-cn.com/problems/find-first-and-last-position-of-element-in-sorted-array/solution/er-fen-cha-zhao-suan-fa-xi-jie-xiang-jie-by-labula/">二分查找算法细节详解，顺便写了首诗</a></p>
]]></content>
</entry>
<entry>
<title></title>
<url>/2023/04/09/LeetCode%E7%AE%97%E6%B3%95%E9%A2%98%E8%A7%A3%E2%80%94%E2%80%94%E5%8F%8C%E6%8C%87%E9%92%882/</url>
<content><![CDATA[<p>@<a href="LeetCode%E7%AE%97%E6%B3%95%E9%A2%98%E8%A7%A3%E2%80%94%E2%80%94%E5%8F%8C%E6%8C%87%E9%92%882">TOC</a></p>
<p>这里介绍双指针在数组中的第二类题型：两端夹击。</p>
<h1 id="第五题"><a href="#第五题" class="headerlink" title="第五题"></a>第五题</h1><p><a href="https://leetcode.cn/problems/squares-of-a-sorted-array/">977. 有序数组的平方</a></p>
<p>题目描述：<br>给你一个按 非递减顺序 排序的整数数组 nums，返回 每个数字的平方 组成的新数组，要求也按 非递减顺序 排序。</p>
<p>示例1:</p>
<figure class="highlight text"><table><tr><td class="code"><pre><span class="line">输入：nums = [-4,-1,0,3,10]</span><br><span class="line">输出：[0,1,9,16,100]</span><br><span class="line">解释：平方后，数组变为 [16,1,0,9,100]</span><br><span class="line">排序后，数组变为 [0,1,9,16,100]</span><br></pre></td></tr></table></figure>
<h2 id="思路"><a href="#思路" class="headerlink" title="思路"></a>思路</h2><p>因为是排序每个数字的平方，根据二次函数图像y=x^2开口向上可得，最大值在两端，最小值在中间。所以，最左和最右进行比较，<strong>两端夹击</strong>，看谁的平方值更大。</p>
<h2 id="代码"><a href="#代码" class="headerlink" title="代码"></a>代码</h2><p>java版本</p>
<figure class="highlight java"><table><tr><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> {</span><br><span class="line"> <span class="keyword">public</span> <span class="type">int</span>[] sortedSquares(<span class="type">int</span>[] nums) {</span><br><span class="line"> <span class="type">int</span> <span class="variable">n</span> <span class="operator">=</span> nums.length;</span><br><span class="line"> <span class="type">int</span>[] result = <span class="keyword">new</span> <span class="title class_">int</span>[n];</span><br><span class="line"> <span class="type">int</span> <span class="variable">l</span> <span class="operator">=</span> <span class="number">0</span>;</span><br><span class="line"> <span class="type">int</span> <span class="variable">r</span> <span class="operator">=</span> n - <span class="number">1</span>;</span><br><span class="line"> <span class="type">int</span> <span class="variable">k</span> <span class="operator">=</span> n - <span class="number">1</span>;</span><br><span class="line"> <span class="keyword">while</span>(l <= r){</span><br><span class="line"> <span class="keyword">if</span>(nums[l] * nums[l] > nums[r] * nums[r]){</span><br><span class="line"> result[k] = nums[l] * nums[l];</span><br><span class="line"> l++;</span><br><span class="line"> k--;</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">else</span>{</span><br><span class="line"> result[k] = nums[r] * nums[r];</span><br><span class="line"> r--;</span><br><span class="line"> k--;</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> result;</span><br><span class="line"> }</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
<p>python版本</p>
<figure class="highlight python"><table><tr><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span>:</span><br><span class="line"> <span class="keyword">def</span> <span class="title function_">sortedSquares</span>(<span class="params">self, nums: <span class="type">List</span>[<span class="built_in">int</span>]</span>) -> <span class="type">List</span>[<span class="built_in">int</span>]:</span><br><span class="line"> n = <span class="built_in">len</span>(nums)</span><br><span class="line"> result = [<span class="number">0</span>] * n</span><br><span class="line"> l, r, idx = <span class="number">0</span>, n - <span class="number">1</span>, n - <span class="number">1</span></span><br><span class="line"> <span class="keyword">while</span> l <= r:</span><br><span class="line"> numl = nums[l] * nums[l]</span><br><span class="line"> numr = nums[r] * nums[r]</span><br><span class="line"> <span class="keyword">if</span> numl > numr:</span><br><span class="line"> result[idx] = numl</span><br><span class="line"> idx -= <span class="number">1</span></span><br><span class="line"> l += <span class="number">1</span></span><br><span class="line"> <span class="keyword">else</span>:</span><br><span class="line"> result[idx] = numr</span><br><span class="line"> idx -= <span class="number">1</span></span><br><span class="line"> r -= <span class="number">1</span></span><br><span class="line"></span><br><span class="line"> <span class="keyword">return</span> result</span><br></pre></td></tr></table></figure>
<h1 id="第六题"><a href="#第六题" class="headerlink" title="第六题"></a>第六题</h1><p><a href="https://leetcode.cn/problems/two-sum-ii-input-array-is-sorted/">167. 两数之和 II - 输入有序数组</a></p>
<p>题目描述:<br>给定一个已按照升序排列的有序数组，找到两个数使得它们相加之和等于目标数。<br>函数应该返回这两个下标值 index1 和 index2，其中 index1 必须小于 index2。</p>
<p>示例1:</p>
<figure class="highlight text"><table><tr><td class="code"><pre><span class="line">输入：numbers = [2,7,11,15], target = 9</span><br><span class="line">输出：[1,2]</span><br><span class="line">解释：2 与 7 之和等于目标数 9 。因此 index1 = 1, index2 = 2 。返回 [1, 2] 。</span><br></pre></td></tr></table></figure>
<h2 id="思路-1"><a href="#思路-1" class="headerlink" title="思路"></a>思路</h2><p>可以先考虑边界情况。升序数组中任意两个元素求和，最小为nums[0]+nums[1]，最大为nums[n-2]+nums[n-1]。target的范围一定在这两者之间，否则找不到答案。所以，我们可以<strong>两端夹击</strong>，一直手抓nums[0]，另一只手抓nums[n-1]。如果是最小的情况，那么让nums[n-1]向nums[1]靠拢；如果是最大的情况，那么让nums[1]向nums[n-2]靠拢。</p>
<h2 id="代码-1"><a href="#代码-1" class="headerlink" title="代码"></a>代码</h2><p>java版本</p>
<figure class="highlight java"><table><tr><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> {</span><br><span class="line"> <span class="keyword">public</span> <span class="type">int</span>[] twoSum(<span class="type">int</span>[] numbers, <span class="type">int</span> target) {</span><br><span class="line"> <span class="type">int</span> <span class="variable">l</span> <span class="operator">=</span> <span class="number">0</span>, r = numbers.length - <span class="number">1</span>;</span><br><span class="line"> <span class="type">int</span>[] result = <span class="keyword">new</span> <span class="title class_">int</span>[]{<span class="number">0</span>, <span class="number">0</span>};</span><br><span class="line"> <span class="keyword">while</span>(l < r){</span><br><span class="line"> <span class="type">int</span> <span class="variable">sum</span> <span class="operator">=</span> numbers[l] + numbers[r];</span><br><span class="line"> <span class="keyword">if</span>(sum == target){</span><br><span class="line"> result[<span class="number">0</span>] = l + <span class="number">1</span>;</span><br><span class="line"> result[<span class="number">1</span>] = r + <span class="number">1</span>;</span><br><span class="line"> <span class="keyword">break</span>;</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">else</span> <span class="keyword">if</span>(sum < target){</span><br><span class="line"> l++;</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">else</span>{</span><br><span class="line"> r--;</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> result;</span><br><span class="line"> }</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
<p>python版本</p>
<figure class="highlight python"><table><tr><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span>:</span><br><span class="line"> <span class="keyword">def</span> <span class="title function_">twoSum</span>(<span class="params">self, numbers: <span class="type">List</span>[<span class="built_in">int</span>], target: <span class="built_in">int</span></span>) -> <span class="type">List</span>[<span class="built_in">int</span>]:</span><br><span class="line"> n = <span class="built_in">len</span>(numbers)</span><br><span class="line"> l, r = <span class="number">0</span>, n - <span class="number">1</span></span><br><span class="line"> <span class="keyword">while</span> l < r:</span><br><span class="line"> m = numbers[l] + numbers[r]</span><br><span class="line"> <span class="keyword">if</span> m == target:</span><br><span class="line"> <span class="keyword">return</span> [l + <span class="number">1</span>, r + <span class="number">1</span>]</span><br><span class="line"> <span class="keyword">elif</span> target > m:</span><br><span class="line"> l += <span class="number">1</span></span><br><span class="line"> <span class="keyword">else</span>:</span><br><span class="line"> r -= <span class="number">1</span></span><br><span class="line"> </span><br><span class="line"> <span class="keyword">return</span> [-<span class="number">1</span>, -<span class="number">1</span>]</span><br></pre></td></tr></table></figure>
<h1 id="第七题"><a href="#第七题" class="headerlink" title="第七题"></a>第七题</h1><p><a href="https://leetcode.cn/problems/sum-of-square-numbers/">633. 平方数之和</a></p>
<p>题目描述:<br>给定一个非负整数 c ，你要判断是否存在两个整数 a 和 b，使得 a^2 + b^2 = c 。</p>
<p>示例1:</p>
<figure class="highlight text"><table><tr><td class="code"><pre><span class="line">输入：c = 5</span><br><span class="line">输出：true</span><br><span class="line">解释：1 * 1 + 2 * 2 = 5</span><br></pre></td></tr></table></figure>
<h2 id="思路-2"><a href="#思路-2" class="headerlink" title="思路"></a>思路</h2><p>上一题是两个数的和是否为target，这道题是两个数的平方和是否为target。一样的思路，考虑最大最小值。若两个数的平方和为target，则最小为0，最大为target的平方根。同样的，<strong>两端夹击</strong>。</p>
<h2 id="代码-2"><a href="#代码-2" class="headerlink" title="代码"></a>代码</h2><p>java版本</p>
<figure class="highlight java"><table><tr><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> {</span><br><span class="line"> <span class="keyword">public</span> <span class="type">boolean</span> <span class="title function_">judgeSquareSum</span><span class="params">(<span class="type">int</span> c)</span> {</span><br><span class="line"> <span class="type">long</span> <span class="variable">l</span> <span class="operator">=</span> <span class="number">0</span>;</span><br><span class="line"> <span class="type">long</span> <span class="variable">r</span> <span class="operator">=</span> (<span class="type">long</span>) Math.sqrt(c);</span><br><span class="line"> <span class="keyword">while</span>(l <= r){</span><br><span class="line"> <span class="type">long</span> <span class="variable">m</span> <span class="operator">=</span> l * l + r * r;</span><br><span class="line"> <span class="keyword">if</span>(m == c){</span><br><span class="line"> <span class="keyword">return</span> <span class="literal">true</span>;</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">else</span> <span class="keyword">if</span>(m < c){</span><br><span class="line"> l++;</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">else</span>{</span><br><span class="line"> r--;</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> <span class="literal">false</span>;</span><br><span class="line"> }</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
<p>python版本</p>
<figure class="highlight python"><table><tr><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span>:</span><br><span class="line"> <span class="keyword">def</span> <span class="title function_">judgeSquareSum</span>(<span class="params">self, c: <span class="built_in">int</span></span>) -> <span class="built_in">bool</span>:</span><br><span class="line"> l, r = <span class="number">0</span>, <span class="built_in">int</span>(sqrt(c))</span><br><span class="line"> <span class="keyword">while</span> l <= r:</span><br><span class="line"> m = l * l + r * r</span><br><span class="line"> <span class="keyword">if</span> m == c:</span><br><span class="line"> <span class="keyword">return</span> <span class="literal">True</span></span><br><span class="line"> <span class="keyword">elif</span> m < c:</span><br><span class="line"> l += <span class="number">1</span></span><br><span class="line"> <span class="keyword">else</span>:</span><br><span class="line"> r -= <span class="number">1</span></span><br><span class="line"> </span><br><span class="line"> <span class="keyword">return</span> <span class="literal">False</span></span><br></pre></td></tr></table></figure>
<p>下一篇博客<a href="">LeetCode算法题解——双指针3</a>中，我将分享LeetCode中关于字符串的双指针问题。</p>
<p>参考：</p>
<p><a href="https://github.com/CyC2018/CS-Notes/blob/master/notes/Leetcode%20%E9%A2%98%E8%A7%A3%20-%20%E5%8F%8C%E6%8C%87%E9%92%88.md">Leetcode 题解 - 双指针</a></p>
]]></content>
</entry>
<entry>
<title></title>
<url>/2023/04/09/Linux%E7%B3%BB%E7%BB%9F%E6%8A%A5%E9%94%99%EF%BC%9AFailed%20to%20execute%20default%20Terminal%20Emulator.%20Input_output%20error./</url>
<content><<br>最近新配了一个服务器环境，之前使用的是配置好anconda和pytorch等库的环境，但是这次申请忘记了之前那个环境是哪一个，所以要重新配置。<br>一登陆之后，打算打开命令行，开始配tensorflow环境，就遇到了一个问题：</p>
<figure class="highlight bash"><table><tr><td class="code"><pre><span class="line">Failed to execute default Terminal Emulator. Input/output error.</span><br></pre></td></tr></table></figure>
<p>命令行打不开了！<br>查询资料之后得知，是Linux系统默认的命令行不对，需要自己手动修改一下：</p>
<ol>
<li>点击桌面左上角的Applications</li>
</ol>
<p><img src="https://img-blog.csdnimg.cn/20200526220342989.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzQxMjE0NjEw,size_5,color_FFFFFF,t_20" alt="在这里插入图片描述"></p>
<ol start="2">
<li>点击Settings，选择Settings Manager<br><img src="https://img-blog.csdnimg.cn/20200526220825366.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzQxMjE0NjEw,size_16,color_FFFFFF,t_70" alt="在这里插入图片描述"></li>
<li>点击Preferred Applications</li>
</ol>
<p><img src="https://img-blog.csdnimg.cn/20200526220935289.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzQxMjE0NjEw,size_16,color_FFFFFF,t_70" alt="在这里插入图片描述"><br>4. 点击Utilities下的Terminal Emulator，选择Xfce Terminal<br><img src="https://img-blog.csdnimg.cn/20200526221002714.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzQxMjE0NjEw,size_16,color_FFFFFF,t_70" alt="在这里插入图片描述"><br>然后就可以正常使用了。<br>参考链接：<a href="https://ubuntuforums.org/showthread.php?t=1894293">Failed to execute default Terminal Emulator. Input/output error.</a></p>
]]></content>
</entry>
<entry>
<title></title>
<url>/2023/04/09/Mac%E4%B8%8B%E5%AE%9E%E7%8E%B0Word%E6%96%87%E6%A1%A3%E6%89%B9%E9%87%8F%E8%BD%AC%E6%8D%A2%E4%B8%BAPDF/</url>
<content><![CDATA[<ol>
<li>使用Adobe Acrobat（首推）<br><a href="https://zhuanlan.zhihu.com/p/89370385">有招啦！轻松实现Word文档批量转换为PDF</a></li>
<li>使用Mac的自动操作<br><a href="https://zhuanlan.zhihu.com/p/391820589">Mac下实现批量Word或PPT转PDF</a></li>
<li>使用LibreOffice<br><a href="https://sspai.com/post/44140#!">如何在 macOS 上一键批量把 PPT 和 Word 文件转成 PDF</a></li>
</ol>
]]></content>
</entry>
<entry>
<title></title>
<url>/2023/04/09/Mac%E4%BD%BF%E7%94%A8ssh%E4%B8%8A%E4%BC%A0%E6%96%87%E4%BB%B6%E5%88%B0%E6%9C%8D%E5%8A%A1%E5%99%A8/</url>
<content><![CDATA[<p>@<a href="Mac%E4%BD%BF%E7%94%A8ssh%E4%B8%8A%E4%BC%A0%E6%96%87%E4%BB%B6%E5%88%B0%E6%9C%8D%E5%8A%A1%E5%99%A8">TOC</a></p>
<figure class="highlight powershell"><table><tr><td class="code"><pre><span class="line">scp <span class="literal">-P</span> 端口号 本地文件 服务器用户名<span class="selector-tag">@</span>服务器ip:目标文件夹</span><br></pre></td></tr></table></figure>
<p>然后，输入”yes”，回车。<br>最后，输入密码，回车。</p>
]]></content>
</entry>
<entry>
<title></title>
<url>/2023/04/09/YOLOv3%E6%98%BE%E7%A4%BA%E7%B1%BB%E5%88%AB%E4%BD%86%E6%98%AF%E4%B8%8D%E6%98%BE%E7%A4%BAbounding%20box/</url>
<content><<br>这两天在跑YOLOv3的源代码，试着用自己的数据集训练并测试。之前测试了好几个数据集都没问题，今天遇到一个数据集，测试结果只显示类别，但是不显示bounding box。而且奇怪的是，并不是所有结果都这样，只是有的类别有，有的类别没有。<br>在原作者GitHub的issues里面找到了答案。<br>原来，bounding box的线条宽度，依赖于输入图片的高度。当输入图片的尺寸太小，bounding box的线条太细，就没法显示了。这个时候需要手动修改src文件夹下image.c/draw_detections函数，将width调大。记得修改文件后，make。</p>
<p><strong>参考</strong><br><a href="https://github.com/pjreddie/darknet/issues/1247">no boxes is shown in predictions.jpg #1247</a></p>
]]></content>
</entry>
<entry>
<title></title>
<url>/2023/04/09/YOLOv3%E8%AE%AD%E7%BB%83+%E4%BF%AE%E6%94%B9/</url>
<content><![CDATA[<p>@<a href="YOLOv3%E8%AE%AD%E7%BB%83+%E4%BF%AE%E6%94%B9">TOC</a><br>最近参加2020年(第13届)中国大学生计算机设计大赛，选择了人工智能挑战赛的赛题二，基于 CT 影像的结直肠息肉检测。<br>赛题要求是，设计算法，判断图像中是否存在息肉，并实现息肉的准确检测，利用矩形方框将所检测出的息肉包含在检测框内。<br>乍一看，是一道标准的目标检测题，再看看官方给出的数据集标注，是YOLO格式的，那就直接用YOLOv3进行训练。最后再根据题目的需求，增加了一些功能。<br>我们这个版本，是在c语言的YOLOv3框架上进行修改的；网上还有很多基于pytorch、tensorflow构建的YOLOv3，以后有时间也写一下。</p>
<h1 id="YOLOv3训练自己的数据集"><a href="#YOLOv3训练自己的数据集" class="headerlink" title="YOLOv3训练自己的数据集"></a>YOLOv3训练自己的数据集</h1><h2 id="标记数据"><a href="#标记数据" class="headerlink" title="标记数据"></a>标记数据</h2><p>如果是训练自己的数据集，即还未对数据集进行标注，那么推荐使用<a href="https://github.com/tzutalin/labelImg">LabelImg</a>工具进行标注，可以得到适用于PascalVOC（xml）或者YOLO（txt）格式的标注。<br>因为官方给出的数据集已经是YOLO格式的标注，我们可以直接用。当然，考虑到后面某个功能使用了PascalVOC格式的标注，这里给出一个从YOLO（txt）格式转换成PascalVOC（xml）格式的代码<a href="https://github.com/LU-K-Brant/yolov3_modify/blob/master/txt2xml.py">txt2xml.py</a>。<br>此时，原始图像放在JPEGImages目录下，YOLO（txt）格式标注放在labels目录下，PascalVOC（xml）格式标注放在Annotations目录下。</p>
<h2 id="制作-yolo-需要的txt文件"><a href="#制作-yolo-需要的txt文件" class="headerlink" title="制作 yolo 需要的txt文件"></a>制作 yolo 需要的txt文件</h2><p>这一步需要制作四个文件：train.txt、val.txt、object_train.txt、object_val.txt。<br>train.txt、val.txt这两个文件保存了用于训练、验证图片的名字，每行一个名字（不带后缀.jpg）。这里参考了一个别人的代码<a href="https://github.com/yehengchen/Object-Detection-and-Tracking/blob/master/OneStage/yolo/yolov3/img2train.py">img2train.py</a>。<br>object_train.txt、object_val.txt这两个文件保存了用于训练、验证图片的绝对路径，每行一个路径。这里参考了一个别人的代码<a href="https://github.com/yehengchen/Object-Detection-and-Tracking/blob/master/OneStage/yolo/yolov3/voc_label.py">voc_label.py</a>。这个代码不仅可以划分文件，还可以将PascalVOC（xml）格式标注转换成YOLO（txt）格式标注。</p>
<h2 id="制作-yolo-需要的配置文件"><a href="#制作-yolo-需要的配置文件" class="headerlink" title="制作 yolo 需要的配置文件"></a>制作 yolo 需要的配置文件</h2><p>这一步需要制作三个文件：ct.names、ct.data、ct.cfg（ct是自己定义的，因为我用的是ct数据集，所以有此定义）。<br>ct.names包含数据集中的种类，每行一个。注意，这个顺序代表了之后预测时的种类顺序。<br>ct.data包含以下几个内容</p>
<blockquote>
<p>classes= 1 # 类别数<br> train = data/object_train.txt # obj_train.txt 路径<br> valid = data/object_val.txt # obj_val.txt 路径<br> names = data/ct.names # ct.names 路径<br> backup = backup/ # 建一个 backup 文件夹用于存放 weights 结果</p>
</blockquote>
<p>注意，这里放的是object_train.txt和object_val.txt，是写有绝对路径的txt文本。<br>ct.cfg包含的是与YOLOv3训练或测试相关的配置，有几个地方需要注意</p>
<blockquote>
<p>1.注意文档开头training和testing的切换；<br>2.直接搜索‘classes’，修改三处对应位置：<br>[convolutional]<br> filters = 3*(classes + 5) #修改filters数量<br> [yolo]<br> classes=5 #修改类别数；<br> 3.修改max_batches = 2000 * classes</p>
</blockquote>
<h2 id="训练"><a href="#训练" class="headerlink" title="训练"></a>训练</h2><p>首先，下载预训练权重。</p>
<figure class="highlight powershell"><table><tr><td class="code"><pre><span class="line"><span class="built_in">wget</span> https://pjreddie.com/media/files/darknet53.conv.<span class="number">74</span></span><br></pre></td></tr></table></figure>
<p>然后，执行训练命令。</p>
<figure class="highlight powershell"><table><tr><td class="code"><pre><span class="line">./darknet detector train ./cfg/ct.data ./cfg/ct.cfg darknet53.conv.<span class="number">74</span></span><br></pre></td></tr></table></figure>
<h1 id="增加功能"><a href="#增加功能" class="headerlink" title="增加功能"></a>增加功能</h1><h2 id="在图像上添加置信度"><a href="#在图像上添加置信度" class="headerlink" title="在图像上添加置信度"></a>在图像上添加置信度</h2><p>YOLOv3单张图像检测的结果，默认设定只包含目标类别。我们可以通过修改src/image.c文件draw_detections()函数，添加目标置信度。修改片段如下：</p>
<figure class="highlight c"><table><tr><td class="code"><pre><span class="line"><span class="keyword">for</span>(i = <span class="number">0</span>; i < num; ++i)</span><br><span class="line">{</span><br><span class="line"> <span class="type">char</span> labelstr[<span class="number">4096</span>] = {<span class="number">0</span>};</span><br><span class="line"> <span class="type">char</span> s1[]={<span class="string">" "</span>};<span class="comment">// 为了name与置信度之间加空格</span></span><br><span class="line"> <span class="type">int</span> <span class="class"><span class="keyword">class</span> =</span> <span class="number">-1</span>;</span><br><span class="line"> <span class="type">char</span> possible[<span class="number">5</span>];<span class="comment">// 存放检测的置信值</span></span><br><span class="line"> <span class="keyword">for</span>(j = <span class="number">0</span>; j < classes; ++j)</span><br><span class="line"> {</span><br><span class="line"> <span class="built_in">sprintf</span>(possible,<span class="string">"%.2f"</span>,dets[i].prob[j]);<span class="comment">//置信值截取小数点后两位赋给possible</span></span><br><span class="line"> <span class="keyword">if</span> (dets[i].prob[j] > thresh)</span><br><span class="line"> {</span><br><span class="line"> <span class="keyword">if</span> (class < <span class="number">0</span>) </span><br><span class="line"> {</span><br><span class="line"> <span class="built_in">strcat</span>(labelstr, names[j]);</span><br><span class="line"> <span class="built_in">strcat</span>(labelstr, s1); <span class="comment">//加空格</span></span><br><span class="line"> <span class="built_in">strcat</span>(labelstr, possible);<span class="comment">//标签中加入置信值</span></span><br><span class="line"> <span class="class"><span class="keyword">class</span> =</span> j;</span><br><span class="line"> } </span><br><span class="line"> <span class="keyword">else</span> </span><br><span class="line"> {</span><br><span class="line"> <span class="built_in">strcat</span>(labelstr, <span class="string">", "</span>);</span><br><span class="line"> <span class="built_in">strcat</span>(labelstr, names[j]);</span><br><span class="line"> <span class="built_in">strcat</span>(labelstr, s1);<span class="comment">//加空格</span></span><br><span class="line"> <span class="built_in">strcat</span>(labelstr, possible);<span class="comment">//标签中加入置信值</span></span><br><span class="line"> }</span><br><span class="line"> <span class="built_in">printf</span>(<span class="string">"%s: %.0f%%\n"</span>, names[j], dets[i].prob[j]*<span class="number">100</span>);</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line">} </span><br></pre></td></tr></table></figure>
<p>单张图像检测命令</p>
<figure class="highlight powershell"><table><tr><td class="code"><pre><span class="line">./darknet detector test ./cfg/ct.data ./cfg/ct_test.cfg ct_final.weights test.jpg</span><br></pre></td></tr></table></figure>
<h2 id="批量检测图像"><a href="#批量检测图像" class="headerlink" title="批量检测图像"></a>批量检测图像</h2><p>首先，修改example/detector.c文件，在开头添加一个获取图片名字的函数：</p>
<figure class="highlight c"><table><tr><td class="code"><pre><span class="line"><span class="type">char</span> *<span class="title function_">GetFilename</span><span class="params">(<span class="type">char</span> *p)</span></span><br><span class="line">{ </span><br><span class="line"> <span class="type">static</span> <span class="type">char</span> name[<span class="number">50</span>]={<span class="string">""</span>};</span><br><span class="line"> <span class="type">char</span> *q = <span class="built_in">strrchr</span>(p,<span class="string">'/'</span>) + <span class="number">1</span>;</span><br><span class="line"> <span class="type">int</span> i = <span class="number">0</span>;</span><br><span class="line"> <span class="keyword">while</span>(q[i] != <span class="string">'\0'</span>)</span><br><span class="line"> {</span><br><span class="line"> <span class="keyword">if</span>(q[i] == <span class="string">'.'</span>) <span class="keyword">break</span>;</span><br><span class="line"> i++;</span><br><span class="line"> }</span><br><span class="line"> <span class="built_in">strncpy</span>(name,q,i);<span class="comment">// i是图片名的长度</span></span><br><span class="line"> name[i] = <span class="string">'\0'</span>;</span><br><span class="line"> <span class="keyword">return</span> name;</span><br><span class="line"></span><br><span class="line">}</span><br></pre></td></tr></table></figure>
<p>在这里，为了对不同长度的文件名能够兼容处理，设置name数组长度为50，可以根据需要修改。<br>然后，替换examples/detector.c 中的test_detector函数：</p>
<figure class="highlight c"><table><tr><td class="code"><pre><span class="line"><span class="type">void</span> <span class="title function_">test_detector</span><span class="params">(<span class="type">char</span> *datacfg, <span class="type">char</span> *cfgfile, <span class="type">char</span> *weightfile, <span class="type">char</span> *filename, <span class="type">float</span> thresh, <span class="type">float</span> hier_thresh, <span class="type">char</span> *outfile, <span class="type">int</span> fullscreen)</span></span><br><span class="line">{</span><br><span class="line"> <span class="built_in">list</span> *options = read_data_cfg(datacfg);</span><br><span class="line"> <span class="type">char</span> *name_list = option_find_str(options, <span class="string">"names"</span>, <span class="string">"data/names.list"</span>);</span><br><span class="line"> <span class="type">char</span> **names = get_labels(name_list);</span><br><span class="line"> </span><br><span class="line"> image **alphabet = load_alphabet();</span><br><span class="line"> network *net = load_network(cfgfile, weightfile, <span class="number">0</span>);</span><br><span class="line"> set_batch_network(net, <span class="number">1</span>);</span><br><span class="line"> srand(<span class="number">2222222</span>);</span><br><span class="line"> <span class="type">double</span> time;</span><br><span class="line"> <span class="type">char</span> buff[<span class="number">256</span>];</span><br><span class="line"> <span class="type">char</span> *input = buff;</span><br><span class="line"> <span class="type">float</span> nms=<span class="number">.45</span>;</span><br><span class="line"> <span class="type">int</span> i=<span class="number">0</span>;</span><br><span class="line"> <span class="keyword">while</span>(<span class="number">1</span>){</span><br><span class="line"> <span class="keyword">if</span>(filename){</span><br><span class="line"> <span class="built_in">strncpy</span>(input, filename, <span class="number">256</span>);</span><br><span class="line"> image im = load_image_color(input,<span class="number">0</span>,<span class="number">0</span>);</span><br><span class="line"> image sized = letterbox_image(im, net->w, net->h);</span><br><span class="line"> layer l = net->layers[net->n<span class="number">-1</span>];</span><br><span class="line"> </span><br><span class="line"> </span><br><span class="line"> <span class="type">float</span> *X = sized.data;</span><br><span class="line"> time=what_time_is_it_now();</span><br><span class="line"> network_predict(net, X);</span><br><span class="line"> <span class="built_in">printf</span>(<span class="string">"%s: Predicted in %f seconds.\n"</span>, input, what_time_is_it_now()-time);</span><br><span class="line"> <span class="type">int</span> nboxes = <span class="number">0</span>;</span><br><span class="line"> detection *dets = get_network_boxes(net, im.w, im.h, thresh, hier_thresh, <span class="number">0</span>, <span class="number">1</span>, &nboxes);</span><br><span class="line"> <span class="keyword">if</span> (nms) do_nms_sort(dets, nboxes, l.classes, nms);</span><br><span class="line"> draw_detections(im, dets, nboxes, thresh, names, alphabet, l.classes);</span><br><span class="line"> free_detections(dets, nboxes);</span><br><span class="line"> <span class="keyword">if</span>(outfile)</span><br><span class="line"> {</span><br><span class="line"> save_image(im, outfile);</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">else</span>{</span><br><span class="line"> save_image(im, <span class="string">"predictions"</span>);</span><br><span class="line"><span class="meta">#<span class="keyword">ifdef</span> OPENCV</span></span><br><span class="line"> cvNamedWindow(<span class="string">"predictions"</span>, CV_WINDOW_NORMAL); </span><br><span class="line"> <span class="keyword">if</span>(fullscreen){</span><br><span class="line"> cvSetWindowProperty(<span class="string">"predictions"</span>, CV_WND_PROP_FULLSCREEN, CV_WINDOW_FULLSCREEN);</span><br><span class="line"> }</span><br><span class="line"> show_image(im, <span class="string">"predictions"</span>,<span class="number">0</span>);</span><br><span class="line"> cvWaitKey(<span class="number">0</span>);</span><br><span class="line"> cvDestroyAllWindows();</span><br><span class="line"><span class="meta">#<span class="keyword">endif</span></span></span><br><span class="line"> }</span><br><span class="line"> free_image(im);</span><br><span class="line"> free_image(sized);</span><br><span class="line"> <span class="keyword">if</span> (filename) <span class="keyword">break</span>;</span><br><span class="line"> } </span><br><span class="line"> <span class="keyword">else</span> {</span><br><span class="line"> <span class="built_in">printf</span>(<span class="string">"Enter Image Path: "</span>);</span><br><span class="line"> fflush(<span class="built_in">stdout</span>);</span><br><span class="line"> input = fgets(input, <span class="number">256</span>, <span class="built_in">stdin</span>);</span><br><span class="line"> <span class="keyword">if</span>(!input) <span class="keyword">return</span>;</span><br><span class="line"> strtok(input, <span class="string">"\n"</span>);</span><br><span class="line"> </span><br><span class="line"> <span class="built_in">list</span> *plist = get_paths(input);</span><br><span class="line"> <span class="type">char</span> **paths = (<span class="type">char</span> **)list_to_array(plist);</span><br><span class="line"> <span class="built_in">printf</span>(<span class="string">"Start Testing!\n"</span>);</span><br><span class="line"> <span class="type">int</span> m = plist->size;</span><br><span class="line"> <span class="keyword">if</span>(access(<span class="string">"/home/lzm/data/test_folder/darknet/car_person/out"</span>,<span class="number">0</span>)==<span class="number">-1</span>)<span class="comment">//"/homelzm/......"修改成自己要保存图片的的路径</span></span><br><span class="line"> {</span><br><span class="line"> <span class="keyword">if</span> (mkdir(<span class="string">"/home/lzm/data/test_folder/darknet/car_person/out"</span>,<span class="number">0777</span>))<span class="comment">//"/homelzm/......"修改成自己要保存图片的的路径</span></span><br><span class="line"> {</span><br><span class="line"> <span class="built_in">printf</span>(<span class="string">"creat folder failed!!!"</span>);</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">for</span>(i = <span class="number">0</span>; i < m; ++i){</span><br><span class="line"> <span class="type">char</span> *path = paths[i];</span><br><span class="line"> image im = load_image_color(path,<span class="number">0</span>,<span class="number">0</span>);</span><br><span class="line"> image sized = letterbox_image(im, net->w, net->h);</span><br><span class="line"> <span class="comment">//image sized = resize_image(im, net->w, net->h);</span></span><br><span class="line"> <span class="comment">//image sized2 = resize_max(im, net->w);</span></span><br><span class="line"> <span class="comment">//image sized = crop_image(sized2, -((net->w - sized2.w)/2), -((net->h - sized2.h)/2), net->w, net->h);</span></span><br><span class="line"> <span class="comment">//resize_network(net, sized.w, sized.h);</span></span><br><span class="line"> layer l = net->layers[net->n<span class="number">-1</span>];</span><br><span class="line"> </span><br><span class="line"> </span><br><span class="line"> <span class="type">float</span> *X = sized.data;</span><br><span class="line"> time=what_time_is_it_now();</span><br><span class="line"> network_predict(net, X);</span><br><span class="line"> <span class="built_in">printf</span>(<span class="string">"Try Very Hard:"</span>);</span><br><span class="line"> <span class="built_in">printf</span>(<span class="string">"%s: Predicted in %f seconds.\n"</span>, path, what_time_is_it_now()-time);</span><br><span class="line"> <span class="type">int</span> nboxes = <span class="number">0</span>;</span><br><span class="line"> detection *dets = get_network_boxes(net, im.w, im.h, thresh, hier_thresh, <span class="number">0</span>, <span class="number">1</span>, &nboxes);</span><br><span class="line"> <span class="comment">//printf("%d\n", nboxes);</span></span><br><span class="line"> <span class="comment">//if (nms) do_nms_obj(boxes, probs, l.w*l.h*l.n, l.classes, nms);</span></span><br><span class="line"> <span class="keyword">if</span> (nms) do_nms_sort(dets, nboxes, l.classes, nms);</span><br><span class="line"> draw_detections(im, dets, nboxes, thresh, names, alphabet, l.classes);</span><br><span class="line"> free_detections(dets, nboxes);</span><br><span class="line"> <span class="keyword">if</span>(outfile){</span><br><span class="line"> save_image(im, outfile);</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">else</span>{</span><br><span class="line"> </span><br><span class="line"> <span class="type">char</span> b[<span class="number">2048</span>];</span><br><span class="line"> <span class="built_in">sprintf</span>(b,<span class="string">"/home/lzm/data/test_folder/darknet/car_person/out/%s"</span>,GetFilename(path));<span class="comment">//"/homelzm/......"修改成自己要保存图片的的路径</span></span><br><span class="line"> </span><br><span class="line"> save_image(im, b);</span><br><span class="line"> <span class="built_in">printf</span>(<span class="string">"OJBK!\n"</span>,GetFilename(path));</span><br><span class="line"><span class="meta">#<span class="keyword">ifdef</span> OPENCV</span></span><br><span class="line"> cvNamedWindow(<span class="string">"predictions"</span>, CV_WINDOW_NORMAL); </span><br><span class="line"> <span class="keyword">if</span>(fullscreen){</span><br><span class="line"> cvSetWindowProperty(<span class="string">"predictions"</span>, CV_WND_PROP_FULLSCREEN, CV_WINDOW_FULLSCREEN);</span><br><span class="line"> }</span><br><span class="line"> show_image(im, <span class="string">"predictions"</span>,<span class="number">0</span>);</span><br><span class="line"> cvWaitKey(<span class="number">0</span>);</span><br><span class="line"> cvDestroyAllWindows();</span><br><span class="line"><span class="meta">#<span class="keyword">endif</span></span></span><br><span class="line"> }</span><br><span class="line"> </span><br><span class="line"> free_image(im);</span><br><span class="line"> free_image(sized);</span><br><span class="line"> <span class="keyword">if</span> (filename) <span class="keyword">break</span>;</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
<p>最后，在命令行输入make，更新文件。<br>批量检测命令，输入的路径为那些图片路径的txt。</p>
<figure class="highlight powershell"><table><tr><td class="code"><pre><span class="line">./darknet detector test ./cfg/ct.data ./cfg/ct_test.cfg ct_final.weights</span><br></pre></td></tr></table></figure>
<h2 id="保存批量检测结果为txt文件"><a href="#保存批量检测结果为txt文件" class="headerlink" title="保存批量检测结果为txt文件"></a>保存批量检测结果为txt文件</h2><p>YOLOv3有自带的命令进行这个操作。</p>
<figure class="highlight powershell"><table><tr><td class="code"><pre><span class="line">./darknet detector valid ./cfg/ct.data ./cfg/ct_test.cfg ct_final.weights results</span><br></pre></td></tr></table></figure>
<h2 id="计算recall"><a href="#计算recall" class="headerlink" title="计算recall"></a>计算recall</h2><p>修改examples/detector.c的validate_detector_recall函数。<br>首先，将validate_detector_recall函数定义和调用修改如下：</p>
<blockquote>
<p>void validate_detector_recall(char *datacfg, char *cfgfile, char *weightfile)<br>validate_detector_recall(datacfg, cfg, weights);</p>
</blockquote>
<p>然后，将如下内容：</p>
<blockquote>
<p>list *plist = get_paths(“data/coco_val_5k.list”);<br>char **paths = (char **)list_to_array(plist);</p>
</blockquote>
<p>修改成</p>
<blockquote>
<p>list *options = read_data_cfg(datacfg);<br>char *valid_images = option_find_str(options, “valid”, “data/train.list”);<br>list *plist = get_paths(valid_images);<br>char **paths = (char **)list_to_array(plist);</p>
</blockquote>
<p>最后，记得make。<br>使用YOLOv3的命令调用方式。</p>
<figure class="highlight powershell"><table><tr><td class="code"><pre><span class="line">./darknet detector recall ./cfg/ct.data ./cfg/ct_test.cfg ct_final.weights</span><br></pre></td></tr></table></figure>
<h2 id="计算mAP"><a href="#计算mAP" class="headerlink" title="计算mAP"></a>计算mAP</h2><p>需要用到PascalVOC（xml）格式的注释，可以用文章开始提到的代码<a href="https://github.com/LU-K-Brant/yolov3_modify/blob/master/txt2xml.py">txt2xml.py</a>进行转换。<br>可以先借助py-faster-rcnn下的<a href="https://github.com/LU-K-Brant/yolov3_modify/blob/master/voc_eval.py">voc_eval.py</a>计算出单个类别的AP，然后求平均得到mAP。<br>新建一个<a href="https://github.com/LU-K-Brant/yolov3_modify/blob/master/all_map.py">all_map.py</a>文件用于计算mAP，这边提供了别的博主的一个例子。<br>如果需要重复计算mAP，需要删除生成的annots.pkl。</p>
<p><strong>参考</strong><br><a href="https://www.cnblogs.com/luzeming/p/10657823.html">YOLOv3：训练自己的数据（附优化与问题总结）</a><br><a href="https://github.com/yehengchen/Object-Detection-and-Tracking/tree/master/OneStage/yolo/yolov3">How to train YOLOv3 model</a><br><a href="https://www.cnblogs.com/xieqi/p/9818056.html">YOLO-V3实战（darknet）</a></p>
]]></content>
</entry>
<entry>
<title></title>
<url>/2023/04/09/LeetCode%E7%AE%97%E6%B3%95%E9%A2%98%E8%A7%A3%E2%80%94%E2%80%94%E9%93%BE%E8%A1%A8%E6%9F%A5%E6%94%B9%E5%A2%9E%E5%88%A0/</url>
<content><![CDATA[<p>@<a href="LeetCode%E7%AE%97%E6%B3%95%E9%A2%98%E8%A7%A3%E2%80%94%E2%80%94%E9%93%BE%E8%A1%A8%E6%9F%A5%E6%94%B9%E5%A2%9E%E5%88%A0">TOC</a></p>
<h1 id="第一题"><a href="#第一题" class="headerlink" title="第一题"></a>第一题</h1><p><a href="https://leetcode.cn/problems/design-linked-list/">707. 设计链表 - 力扣（LeetCode）</a></p>
<h3 id="代码"><a href="#代码" class="headerlink" title="代码"></a>代码</h3><p>python版本</p>
<figure class="highlight python"><table><tr><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">MyLinkedList</span>:</span><br><span class="line"></span><br><span class="line"> <span class="keyword">def</span> <span class="title function_">__init__</span>(<span class="params">self</span>):</span><br><span class="line"> self.dummy = ListNode(-<span class="number">1</span>)</span><br><span class="line"> self.size = <span class="number">0</span></span><br><span class="line"></span><br><span class="line"> <span class="keyword">def</span> <span class="title function_">get</span>(<span class="params">self, index: <span class="built_in">int</span></span>) -> <span class="built_in">int</span>:</span><br><span class="line"> <span class="keyword">if</span> index < <span class="number">0</span> <span class="keyword">or</span> index >= self.size:</span><br><span class="line"> <span class="keyword">return</span> -<span class="number">1</span></span><br><span class="line"></span><br><span class="line"> cur = self.dummy.<span class="built_in">next</span></span><br><span class="line"> i = <span class="number">0</span></span><br><span class="line"> <span class="keyword">while</span> i < index:</span><br><span class="line"> cur = cur.<span class="built_in">next</span></span><br><span class="line"> i += <span class="number">1</span></span><br><span class="line"></span><br><span class="line"> <span class="keyword">return</span> cur.val</span><br><span class="line"></span><br><span class="line"> <span class="keyword">def</span> <span class="title function_">addAtHead</span>(<span class="params">self, val: <span class="built_in">int</span></span>) -> <span class="literal">None</span>:</span><br><span class="line"> self.addAtIndex(<span class="number">0</span>, val)</span><br><span class="line"></span><br><span class="line"> <span class="keyword">def</span> <span class="title function_">addAtTail</span>(<span class="params">self, val: <span class="built_in">int</span></span>) -> <span class="literal">None</span>:</span><br><span class="line"> self.addAtIndex(self.size, val)</span><br><span class="line"></span><br><span class="line"> <span class="keyword">def</span> <span class="title function_">addAtIndex</span>(<span class="params">self, index: <span class="built_in">int</span>, val: <span class="built_in">int</span></span>) -> <span class="literal">None</span>:</span><br><span class="line"> <span class="keyword">if</span> index < <span class="number">0</span>:</span><br><span class="line"> index = <span class="number">0</span></span><br><span class="line"> <span class="keyword">elif</span> index > self.size:</span><br><span class="line"> <span class="keyword">return</span></span><br><span class="line"></span><br><span class="line"> pre = self.dummy</span><br><span class="line"> i = <span class="number">0</span></span><br><span class="line"> <span class="keyword">while</span> i < index:</span><br><span class="line"> pre = pre.<span class="built_in">next</span></span><br><span class="line"> i += <span class="number">1</span></span><br><span class="line"></span><br><span class="line"> pre.<span class="built_in">next</span> = ListNode(val, pre.<span class="built_in">next</span>)</span><br><span class="line"> self.size += <span class="number">1</span></span><br><span class="line"> <span class="keyword">return</span></span><br><span class="line"></span><br><span class="line"> <span class="keyword">def</span> <span class="title function_">deleteAtIndex</span>(<span class="params">self, index: <span class="built_in">int</span></span>) -> <span class="literal">None</span>:</span><br><span class="line"> <span class="keyword">if</span> index < <span class="number">0</span> <span class="keyword">or</span> index >= self.size:</span><br><span class="line"> <span class="keyword">return</span></span><br><span class="line"></span><br><span class="line"> pre = self.dummy</span><br><span class="line"> i = <span class="number">0</span></span><br><span class="line"> <span class="keyword">while</span> i < index:</span><br><span class="line"> pre = pre.<span class="built_in">next</span></span><br><span class="line"> i += <span class="number">1</span></span><br><span class="line"></span><br><span class="line"> pre.<span class="built_in">next</span> = pre.<span class="built_in">next</span>.<span class="built_in">next</span></span><br><span class="line"> self.size -= <span class="number">1</span></span><br><span class="line"> <span class="keyword">return</span></span><br></pre></td></tr></table></figure>
<h1 id="第二题"><a href="#第二题" class="headerlink" title="第二题"></a>第二题</h1><p><a href="https://leetcode.cn/problems/middle-of-the-linked-list/">876. 链表的中间结点 - 力扣（LeetCode）</a></p>
<h3 id="代码-1"><a href="#代码-1" class="headerlink" title="代码"></a>代码</h3><p>python版本</p>
<figure class="highlight python"><table><tr><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span>:</span><br><span class="line"> <span class="keyword">def</span> <span class="title function_">middleNode</span>(<span class="params">self, head: <span class="type">Optional</span>[ListNode]</span>) -> <span class="type">Optional</span>[ListNode]:</span><br><span class="line"> slow = fast = head</span><br><span class="line"> <span class="keyword">while</span> fast <span class="keyword">and</span> fast.<span class="built_in">next</span>:</span><br><span class="line"> fast = fast.<span class="built_in">next</span>.<span class="built_in">next</span></span><br><span class="line"> slow = slow.<span class="built_in">next</span></span><br><span class="line"></span><br><span class="line"> <span class="keyword">return</span> slow</span><br></pre></td></tr></table></figure>
<h1 id="第三题"><a href="#第三题" class="headerlink" title="第三题"></a>第三题</h1><p><a href="https://leetcode.cn/problems/remove-linked-list-elements/">203. 移除链表元素 - 力扣（LeetCode）</a></p>
<h3 id="代码-2"><a href="#代码-2" class="headerlink" title="代码"></a>代码</h3><p>python版本</p>
<figure class="highlight python"><table><tr><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span>:</span><br><span class="line"> <span class="keyword">def</span> <span class="title function_">removeElements</span>(<span class="params">self, head: <span class="type">Optional</span>[ListNode], val: <span class="built_in">int</span></span>) -> <span class="type">Optional</span>[ListNode]:</span><br><span class="line"> dummy = ListNode(-<span class="number">1</span>, head)</span><br><span class="line"> cur = dummy</span><br><span class="line"> <span class="keyword">while</span> cur.<span class="built_in">next</span>:</span><br><span class="line"> <span class="keyword">if</span> cur.<span class="built_in">next</span>.val == val:</span><br><span class="line"> cur.<span class="built_in">next</span> = cur.<span class="built_in">next</span>.<span class="built_in">next</span></span><br><span class="line"> <span class="keyword">else</span>:</span><br><span class="line"> cur = cur.<span class="built_in">next</span></span><br><span class="line"> <span class="keyword">return</span> dummy.<span class="built_in">next</span></span><br></pre></td></tr></table></figure>
<h1 id="第四题"><a href="#第四题" class="headerlink" title="第四题"></a>第四题</h1><p><a href="https://leetcode.cn/problems/remove-duplicates-from-sorted-list/">83. 删除排序链表中的重复元素 - 力扣（LeetCode）</a></p>
<h3 id="代码-3"><a href="#代码-3" class="headerlink" title="代码"></a>代码</h3><p>python版本</p>
<figure class="highlight python"><table><tr><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span>:</span><br><span class="line"> <span class="keyword">def</span> <span class="title function_">deleteDuplicates</span>(<span class="params">self, head: <span class="type">Optional</span>[ListNode]</span>) -> <span class="type">Optional</span>[ListNode]:</span><br><span class="line"> dummy = ListNode(-<span class="number">101</span>, head)</span><br><span class="line"> cur = dummy</span><br><span class="line"> <span class="keyword">while</span> cur.<span class="built_in">next</span>:</span><br><span class="line"> <span class="keyword">if</span> cur.val == cur.<span class="built_in">next</span>.val:</span><br><span class="line"> cur.<span class="built_in">next</span> = cur.<span class="built_in">next</span>.<span class="built_in">next</span></span><br><span class="line"> <span class="keyword">else</span>:</span><br><span class="line"> cur = cur.<span class="built_in">next</span></span><br><span class="line"> </span><br><span class="line"> <span class="keyword">return</span> dummy.<span class="built_in">next</span></span><br></pre></td></tr></table></figure>
<h1 id="第五题"><a href="#第五题" class="headerlink" title="第五题"></a>第五题</h1><p><a href="https://leetcode.cn/problems/remove-nth-node-from-end-of-list/">19. 删除链表的倒数第 N 个结点 - 力扣（LeetCode）</a></p>
<h3 id="代码-4"><a href="#代码-4" class="headerlink" title="代码"></a>代码</h3><p>python版本</p>
<figure class="highlight python"><table><tr><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span>:</span><br><span class="line"> <span class="keyword">def</span> <span class="title function_">removeNthFromEnd</span>(<span class="params">self, head: <span class="type">Optional</span>[ListNode], n: <span class="built_in">int</span></span>) -> <span class="type">Optional</span>[ListNode]:</span><br><span class="line"> dummy = ListNode(-<span class="number">1</span>, head)</span><br><span class="line"> slow = fast = dummy</span><br><span class="line"></span><br><span class="line"> i = <span class="number">0</span></span><br><span class="line"> <span class="keyword">while</span> i < n:</span><br><span class="line"> fast = fast.<span class="built_in">next</span></span><br><span class="line"> i += <span class="number">1</span></span><br><span class="line"></span><br><span class="line"> <span class="keyword">while</span> fast.<span class="built_in">next</span>:</span><br><span class="line"> fast = fast.<span class="built_in">next</span></span><br><span class="line"> slow = slow.<span class="built_in">next</span></span><br><span class="line"></span><br><span class="line"> slow.<span class="built_in">next</span> = slow.<span class="built_in">next</span>.<span class="built_in">next</span></span><br><span class="line"> <span class="keyword">return</span> dummy.<span class="built_in">next</span></span><br></pre></td></tr></table></figure>
]]></content>
</entry>
<entry>
<title>“How the economic machine works”观后感</title>
<url>/2021/04/25/how%20the%20economic%20machine%20works/</url>
<content><![CDATA[<h1 id="摘要"><a href="#摘要" class="headerlink" title="摘要"></a>摘要</h1><p>不同于传统复杂的经济模型，Ray Dalio总结出一个简单易懂的经济模型，包括经济运行的三种规模、四类个体和三股动力。此外，Ray Dalio还给出避免陷入经济困境的三条经验法则。</p>
<span id="more"></span>
<h1 id="引言"><a href="#引言" class="headerlink" title="引言"></a>引言</h1><p>大家都听过一句话，经济基础决定上层建筑。<!--这一句过渡有点不自然。-->那么这里的“经济”是如何运行的呢？有很多经济学著作已经给出了答案，无论是西方经济学，抑或是社会主义经济学。<!--这里举几个例子。-->但是这些经济学著作，无一不长篇大论，让人一时难以摸到门槛。那么能不能用一个简单易懂的模型，来描述经济的运行呢？</p>
<p>在视频“How the economic machine works”中，Ray Dalio用30分钟描绘出一个简单易懂的经济模型。<!--Ray Dalio，2020年福布斯富豪排行榜第46位，身价180亿美元。-->在Ray Dalio的经济模型中，根据对经济运行起作用的时间范围，经济运行可以看作由三股动力支持：生产率、短期债务周期和长期债务周期。为了更细致地介绍，Ray Dalio将经济运行分为三种规模：交易、市场和经济。按照在经济中所起的作用不同，经济中的主体可以分成四类：个体、企业、银行和政府。在三股动力的共同影响下，不同主体在各种规模的经济活动中作出反应，经济机器便不停地运行起来。</p>
<h1 id="Ray-Dalio的经济模型"><a href="#Ray-Dalio的经济模型" class="headerlink" title="Ray Dalio的经济模型"></a>Ray Dalio的经济模型</h1><h2 id="经济运行的三种规模"><a href="#经济运行的三种规模" class="headerlink" title="经济运行的三种规模"></a>经济运行的三种规模</h2><h3 id="交易"><a href="#交易" class="headerlink" title="交易"></a>交易</h3><p>我们先来假设最简单的一种经济运行——只有买方和卖方两个人参与的经济运行。买方提供货币或信贷，卖方提供商品、服务或金融资产，如此便完成了一次交易。这里，货币和信贷也称为支出总额，商品、服务和金融资产称为产销总量，支出总额/产销总量=价格。一个人走进了一间酒吧，要了一杯啤酒，留下了十块钱，那这个人和酒吧老板就完成了一次交易。可见，交易是规模最小、情况最简单的经济运行。</p>
<h3 id="市场"><a href="#市场" class="headerlink" title="市场"></a>市场</h3><p>当交易某一种商品（广义，包括服务和金融资产）的所有买方和卖方聚集在一起，就形成了市场。例如，炎炎夏日的一个公园里，所有卖冰淇淋的商贩和买冰淇淋的小屁孩，就构成了一个小小的冰淇淋市场。<!--如果没有垄断的话，那他们就是一个自由市场。--></p>
<h3 id="经济"><a href="#经济" class="headerlink" title="经济"></a>经济</h3><p>当我们把所有市场中的全部交易都考虑进来，那么我们所谈论的就是经济。</p>
<h2 id="经济运行的四类个体"><a href="#经济运行的四类个体" class="headerlink" title="经济运行的四类个体"></a>经济运行的四类个体</h2><p>按照在经济中所起的作用不同，经济中的主体大致可以分成四类：个体、企业、银行和政府。</p>
<h3 id="个体"><a href="#个体" class="headerlink" title="个体"></a>个体</h3><p>每一个普通人，都是经济运行中的个体。正如前文所提到的，买啤酒的人和买冰淇淋的小朋友，他们是经济运行中的个体。</p>
<h3 id="企业"><a href="#企业" class="headerlink" title="企业"></a>企业</h3><p>企业的职能是组织工人进行生产和销售。</p>
<h3 id="银行"><a href="#银行" class="headerlink" title="银行"></a>银行</h3><p>银行的职能是储蓄和贷款。</p>
<h3 id="政府"><a href="#政府" class="headerlink" title="政府"></a>政府</h3><p>在这里，Ray Dalio将政府又进一步细分成了中央政府和中央银行。中央政府的职能是收税、发行国债和发放救助金。中央银行的职能是调控利率和印发钞票。</p>
<h2 id="经济运行的三股动力"><a href="#经济运行的三股动力" class="headerlink" title="经济运行的三股动力"></a>经济运行的三股动力</h2><h3 id="生产率"><a href="#生产率" class="headerlink" title="生产率"></a>生产率</h3><p>让我们再次将目光聚焦于一场交易中。</p>
<p>我们先考虑交易中的卖方。我们假定卖方所出售的商品由卖方生产，那么商品的数量多少与质量好坏，只取决于卖方的生产率。如果卖方的生产率高，那么卖方就能生产出高于市场平均水平的商品，也就能获得高于市场平均水平的收入。</p>
<p>我们再来考虑交易中的买方。我们假定买方所用于购买商品的货币只来自收入（这里不考虑信贷），那么参考上面卖方的情况我们可以得知，买方可用于购买商品的货币多少，也只取决于买方的生产率。如果买方的生产率高，那么买方就能获得高于市场平均水平的收入，也就能购买高于市场平均水平的商品。</p>
<p>综上所述，在这种情况下，一个人的支出只取决于一个人的收入，而一个人的收入又只取决于他的生产率。要想提高生产率，就只有不断改进生产工艺、提高生产效率，这需要长时间的积累沉淀。所以，生产率是推动经济运行最长期、最根本的动力。</p>
<h3 id="短期债务周期"><a href="#短期债务周期" class="headerlink" title="短期债务周期"></a>短期债务周期</h3><p>我们现在考虑信贷对交易双方的影响。</p>
<p>在交易中的买方，通过信贷的方式获得了更多的货币。现在他能够购买商品的支出=收入+信贷，也就是他的支出能够大于收入。一个人的支出是另一个人的收入，买方的支出增加会让卖方的收入增加。支出-收入的链式法则不断传播，信贷的放大效应继续扩大，交易活动变得越来越多。在短时间内，有信贷的经济运行在只依赖生产率提高的经济运行更加活泼。</p>
<p>为什么是短时间内呢？因为出来混，总是要还的。在这个阶段你通过信贷的方式，使得支出能够超过收入，那么在下一个阶段偿还信贷时，你的支出必定会小于收入。在这一借一还之间，短期债务周期就产生了。</p>
<p>我们再来看看信贷对整个经济的影响。</p>
<p>我们之前说过，支出总额是由货币和信贷构成的。当信贷增加，那么整个市场的支出总额增加，在产销总量跟不上的情况下（要知道，生产出一个实实在在的商品可是要花时间的，短时间内不可能产量爆炸增长），价格就会飞升，我们称之为通货膨胀。此时就需要中央银行出手，上调利率，使得借债成本上升，那么信贷的数量就会下降，整个市场中流通的钱又会减少，价格回到正常的位置。</p>
<p>反过来，如果信贷减少，整个市场的支出总额减少，而产销总量却在上升（商品生产的变化可是比较慢的哟，不能说增产就增产，说减产就减产），那么价格就会骤降，我们称之为通货紧缩。此时又到了中央银行出手的时候，下调利率，使得借债成本下降，那么信贷的数量就会上升，整个市场中流通的钱又会增加，价格回到正常的位置。</p>
<p>在短期债务周期中，整个过程主要由中央银行控制，通过控制利率来调控信贷数量，经济增长的速度取决于信贷数量的多少。</p>
<p>既然如此，通胀的时候上调利率，通缩的时候下调利率，为什么还会有长期债务周期呢？这是由人的天性决定的。当一个人开始借钱之后，他就会倾向于借更多的钱，久而久之就会积累出难以偿还的债务，也就形成了长期债务周期。</p>
<p>但是请注意，这并不是说信贷不好。在之前已经说了，相比较只依赖生产率提高的经济运行，信贷能够在一段时间内促进经济发展。特别是。当你将信贷用于购买促进生产率提高的商品时，你的信贷会给你带来收入，使得你有能力偿还信贷，那么这一切就是良性的。可是，如果你将信贷用于购买满足消费欲望的商品，并且不断举债以丰富物质生活，那么你必然要面对难以承受的后果。</p>
<h3 id="长期债务周期"><a href="#长期债务周期" class="headerlink" title="长期债务周期"></a>长期债务周期</h3><p>人们在不断的借债过程中，发现用借来的钱进行投资是很不错的选择，于是金融资产的价格不断攀升，由此也就产生了“泡沫”。越来越多的人开始借越来越多的钱，越来越多的钱开始涌入越来越多的金融资产，越来越多的金融资产开始变得越来越值钱，世界经济一片欣欣向荣。</p>
<p>这一切都很美好，不是吗？是的，如果借钱不用还的话，这一切的确是很美好！</p>
<p>当人们借的钱越来越多，已经到了没办法拆东墙补西墙的时候，开始减少支出、出售金融资产以偿还债务。这里再一次强调，一个人的支出是另一个人的收入。当人们减少支出，经济活动开始减弱；当人们减少购买甚至出售金融资产时，金融资产的价格就会不断下跌。而当所有的个体、企业、银行和政府，都背负着极高的债务需要偿还时，经济萧条就在眼前。</p>
<p>那么我们可不可以采取在短期债务周期中使用过的办法，通过中央银行下调利率，让信贷增加，以此促进经济呢？答案是，不可以。因为在经历过多个短期债务周期后，利率已经处于极低的水平（不然怎么大家借的钱越来越多呢？）；而且极高的债务负担削弱了借贷的欲望（有钱的不敢借出，没钱的也不敢借入，大家都怕还不上钱）。这个时候，我们就需要其他手段解决问题，我们称之为“去杠杆化”。</p>
<p>“去杠杆化”的办法总的来说有四种：削减支出、减少债务、财富再分配和发行货币。</p>
<p>削减支出，让收入尽可能多地用来偿还债务，那么不就可以减轻债务负担了吗？这个听上去很不错，但是不要忘记了，一个人的支出是另一个人的收入。当整个社会都背负高债务负担时，如果大家都选择削减支出，那么经济活动就会大量减少，市场上流通的钱减少，通货紧缩又一次形成了。</p>
<p>减少债务，让还款的时间延迟，或下调还款的利息，能不能解决问题呢？听上去很有道理，既然大家背负很重的债务负担还不起，那么大家能不能妥协一下，迟还或者少还一点。这里又遇到了和“削减支出”类似的问题，信贷是支出的重要组成部分，当信贷还款减少了，就意味着有人的收入减少了，市场上的流通的钱减少，通货紧缩又一次形成了。</p>
<p>既然问题出在钱减少了，那么我们找出更多的钱不就可以了？谁有更多的钱呢，很自然的，有钱人会有更多的钱。财富再分配，就是指中央政府（想想中央政府的职能）向有钱人征收更多的税，用以发放救助金和开展基建项目。可是当整个社会背负极高的债务负担时，地主家也没有多少余粮啊。有钱的企业家们，不仅要面对经济下行带来的影响（生意不好做），还要被中央政府收一笔税，他们的日子也不好过，经济会继续下行，带来通货紧缩。</p>
<p>哪里还能找到更多的钱呢，或者说，如何产生更多的钱？中央银行的一项重要职能就是印发钞票。只要让中央银行开动印钞机，就能源源不断地产生钱啦。用中央银行印出来的钞票，购买中央政府发行的国债（所以货币也可以看作是政府的信用），相当于把多出来的钱交到了中央政府手里，那么中央政府就有钱发放救助金和开展基建项目。就这样，凭空产生的钱就送到了大家的手上，大家可以用来偿还债务和购买商品。但是货币增长过快，超出了商品增长的速度，价格就会飞涨，带来通货膨胀。</p>
<p>“去杠杆化”的四种办法，前三种办法会导致通货紧缩，第四种办法会带来通货膨胀。只要我们能够合理利用这四种办法，平衡通缩和通胀，就能实现和谐的“去杠杆化”。“去杠杆化”的核心在于，要让收入的增长大于债务的增长，从而消化掉整个社会的高负债。在这个过程中，既不能形成严重的通胀，也不能形成严重的通缩。</p>
<p>纵观整个长期债务周期，“杠杆化”的时间大概是50年左右，其中由多个短期债务周期构成，整个社会都在不断举债消费；而萧条期大概持续2至3年，长期积累的债务危机爆发了；要想消化掉这一波债务，恢复到以前的经济水平，大概需要7到10年的“再膨胀”。从萧条到再膨胀，就是人们常说的“失去的十年”。</p>
<h1 id="总结"><a href="#总结" class="headerlink" title="总结"></a>总结</h1><p>在最后，Ray Dalio给出了自己应对经济运行的三条经验法则。</p>
<p>经验一，不要让债务增长超过收入增长，否则无法偿还的债务会压垮你。</p>
<p>经验二，不要让收入增长超过生产率增长，否则你会在工作中失去竞争力。</p>
<p>经验三，尽一切努力提高生产率，因为长期来看生产率起决定作用。</p>
]]></content>
<categories>
<category>观/读后感</category>
</categories>
<tags>
<tag>经济</tag>
<tag>Ray Dalio</tag>
</tags>
</entry>
<entry>
<title></title>
<url>/2023/04/09/mac%E6%89%93%E5%BC%80%E7%BD%91%E9%A1%B5%E9%80%9F%E5%BA%A6%E7%89%B9%E5%88%AB%E6%85%A2/</url>
<content><![CDATA[<p><strong>问题描述</strong>：用MacBook Pro笔记本电脑，确定连接了网络，其他应用上网也没问题，就是浏览器打开网页的速度特别慢。换各种浏览器safari，firefox，chorme等打开网页都很慢。<br><strong>解决方案</strong>：不是电脑硬件的问题，不是网络连接的问题，那应该就是网络配置的问题了。<br><img src="https://img-blog.csdnimg.cn/20201026194145448.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzQxMjE0NjEw,size_16,color_FFFFFF,t_70#pic_center" alt="系统偏好设置-网络"></p>
<p>打开系统偏好设置，点击网络，在上方的“位置”，选择“编辑位置”，编辑一个“家”。立马解决问题。</p>
<p><strong>参考</strong>：<br><a href="https://www.zhihu.com/question/24732601">mac打开网页速度特别慢？</a></p>
]]></content>
</entry>
<entry>
<title>全国第四轮学科评估结果查询</title>
<url>/2023/04/09/%E5%85%A8%E5%9B%BD%E7%AC%AC%E5%9B%9B%E8%BD%AE%E5%AD%A6%E7%A7%91%E8%AF%84%E4%BC%B0%E7%BB%93%E6%9E%9C%E6%9F%A5%E8%AF%A2/</url>
<content><![CDATA[<p>最近出了高考成绩，有些学弟学妹找我咨询志愿填报的问题。<br>无论是“地域优先论”，还是“学校优先论”，或者是“专业优先论”，我都持平等的看法。<br>读大学，说到底还是对一个人基本素质的培养，其他都是虚的。不过，为了满足学弟学妹们的需要，我也整理了一些资料，提供给大家。</p>
<span id="more"></span>
<p>分享一个<a href="https://souky.eol.cn/api/newapi/assess_result#">全国第四轮学科评估结果查询网址</a>。这个网址可以分专业查看不同大学的排名，或者分大学查看不同专业的排名。</p>
<p>其他资料，收集了再补充。</p>
]]></content>
<categories>
<category>杂谈</category>
</categories>
</entry>
<entry>
<title>为GitHub的私有项目添加合作者</title>
<url>/2023/04/09/%E4%B8%BAGitHub%E7%9A%84%E7%A7%81%E6%9C%89(Private)%E9%A1%B9%E7%9B%AE%E6%B7%BB%E5%8A%A0%E5%90%88%E4%BD%9C%E8%80%85(collaborator)/</url>
<content><![CDATA[<p>当你在GitHub中当仓库设置为private时，对外是不可见的，对方无法从GitHub主页查找你，也无法通过你分享的网址查找你。<br>想要让其他人加入到这个私人项目中，可以添加合作者。</p>
<span id="more"></span>
<p>首先，进入你的私人仓库，找到Settings。<br><img src="https://img-blog.csdnimg.cn/20200731110059523.png" alt="在这里插入图片描述"><br>然后，点击下方左侧的Manage access。<br><img src="https://img-blog.csdnimg.cn/20200731110328741.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzQxMjE0NjEw,size_16,color_FFFFFF,t_70" alt="在这里插入图片描述"><br>最后，选择右侧的Invite a collaborator。<br><img src="https://img-blog.csdnimg.cn/20200731110624119.png" alt="在这里插入图片描述"><br>添加对方的GitHub账号，然后确定。将邀请链接发给对方，对方打开同意即可。<br><img src="https://img-blog.csdnimg.cn/20200731111017794.png" alt="在这里插入图片描述"><br>合作者可以在Setting的Repositories最底部看到这个仓库。<br><img src="https://img-blog.csdnimg.cn/2020073111240098.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzQxMjE0NjEw,size_16,color_FFFFFF,t_70" alt="在这里插入图片描述"><br><img src="https://img-blog.csdnimg.cn/20200731112707501.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzQxMjE0NjEw,size_16,color_FFFFFF,t_70" alt="在这里插入图片描述"><br><strong>参考</strong><br><a href="https://blog.csdn.net/qq_35197701/article/details/106672583">github私有库授权</a><br><a href="https://blog.csdn.net/chenbetter1996/article/details/82871518">GitHub 私人private仓库添加成员（协作者Collaborators）</a></p>
]]></content>
<categories>
<category>计算机技术</category>
</categories>
</entry>
<entry>
<title>如何在word中插入LaTeX公式</title>
<url>/2021/04/29/%E5%A6%82%E4%BD%95%E5%9C%A8word%E4%B8%AD%E6%8F%92%E5%85%A5LaTeX%E5%85%AC%E5%BC%8F/</url>
<content><![CDATA[<h1 id="摘要"><a href="#摘要" class="headerlink" title="摘要"></a>摘要</h1><p>很多时候会遇到用LaTeX写好论文，然后要改写到Word中的情况，这个时候公式的处理是一个比较麻烦的事情。有三种办法可以简便地将LaTeX中的公式代码插入到Word中。第一种办法最方便，使用MathType，将LaTeX转成Word公式，可惜要收费。第二种是使用Word2019，可以直接插入LaTeX格式的公式；第三种是将LaTeX转成MathML格式插入Word。</p>
<span id="more"></span>
<h1 id="使用MathType"><a href="#使用MathType" class="headerlink" title="使用MathType"></a>使用MathType</h1><p>因为MathType只给了30天免试用，之后要收费，所以在这里不详细介绍，给出一个官网文章<a href="#refer-anchor-6"><sup>6</sup></a>作为参考。</p>
<h1 id="使用Word2019"><a href="#使用Word2019" class="headerlink" title="使用Word2019"></a>使用Word2019</h1><p>使用Word2019直接插入LaTeX格式的公式。<a href="#refer-anchor-1"><sup>1</sup></a><sup>,</sup><a href="#refer-anchor-2"><sup>2</sup></a></p>
<ol>
<li><p>在Word2019中新建一个文件，Windows系统下按“alt”+“=”插入公式编辑框，macOS系统下按“control”+“=”插入公式编辑框。</p>
</li>
<li><p>在菜单栏选择“插入”、“公式”，选择“{}LaTeX”。</p>
</li>
<li><p>将编辑好的LaTeX代码粘贴到公式编辑框中，按“回车键”。</p>
</li>
<li><p>选中公式，右键，选择“段落”，调整“行距”为“单倍行距”。<a href="#refer-anchor-3"><sup>3</sup></a></p>
</li>
</ol>
<h1 id="使用Mathpix-Snip或latexlive"><a href="#使用Mathpix-Snip或latexlive" class="headerlink" title="使用Mathpix Snip或latexlive"></a>使用Mathpix Snip或latexlive</h1><p>然后我们讲讲第二种方法，如何将LaTeX转成MathML格式的公式插入Word。</p>
<ol>
<li>使用Mathpix Snip<a href="#refer-anchor-4"><sup>4</sup></a>或latexlive<a href="#refer-anchor-5"><sup>5</sup></a>将LaTeX代码转成MathML格式。</li>
<li>在Word2019中新建一个文件，Windows系统下按“alt”+“=”插入公式编辑框，macOS系统下按“control”+“=”插入公式编辑框。</li>
<li>复制MathML格式的代码。以<strong>纯文本</strong>的格式粘贴到公式编辑框中，按“回车键”。</li>
<li>选中公式，右键，选择“段落”，调整“行距”为“单倍行距”。<a href="#refer-anchor-3"><sup>3</sup></a></li>
</ol>
<h2 id="参考"><a href="#参考" class="headerlink" title="参考"></a>参考</h2><div id="refer-anchor-1"></div>
<p>[1] <a href="https://cloud.tencent.com/developer/article/1560570">Word2019插入LaTex编码的公式</a></p>
<div id="refer-anchor-2"></div>
<p>[2] <a href="https://www.jianshu.com/p/47152773e53c">如何在word中使用latex语言优雅的编辑数学公式？</a></p>
<div id="refer-anchor-3"></div>
<p>[3] <a href="https://www.jianshu.com/p/17ef28b1be49">在Word中优雅地插入Latex公式</a></p>
<div id="refer-anchor-4"></div>
<p>[4] <a href="https://zhuanlan.zhihu.com/p/137641360">Mathpix Snip 截图识别公式插入 Microsoft Word 中——无需mathtype</a></p>
<div id="refer-anchor-5"></div>
<p>[5] <a href="https://blog.csdn.net/weixin_45026882/article/details/107342524">【LaTeX】公式书写工具 | LaTeX转MathML（Word适用）| 公式截图转LaTeX| LaTeX学习（偏公式输入方面）</a></p>
<div id="refer-anchor-6"></div>
<p>[6] <a href="https://www.mathtype.cn/jiqiao/daima-bianji.html">如何在MathType中用LaTex代码编辑公式</a></p>
]]></content>
<categories>
<category>计算机技术</category>
</categories>
<tags>
<tag>论文写作</tag>
<tag>Word</tag>
<tag>LaTeX</tag>
</tags>
</entry>
<entry>
<title></title>
<url>/2023/04/09/linux%E4%B8%8Btensorflow%E4%B8%8Ecuda%E7%89%88%E6%9C%AC%E4%B8%8D%E5%85%BC%E5%AE%B9%E9%97%AE%E9%A2%98%EF%BC%8CImportError_%20libcublas.so.10.0_%20cannot%20open%20shared%20object%20file/</url>
<content><![CDATA[<p>@<a href="linux%E4%B8%8Btensorflow%E4%B8%8Ecuda%E7%89%88%E6%9C%AC%E4%B8%8D%E5%85%BC%E5%AE%B9%E9%97%AE%E9%A2%98">TOC</a><br>因为跑一个多目标跟踪的代码，要用到yolov3，所以作为pytorch用户的我，第一次用上了tensorflow。<br>但是我使用的镜像环境，是cuda10.1版本的：<br><img src="https://img-blog.csdnimg.cn/20200526222813753.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzQxMjE0NjEw,size_16,color_FFFFFF,t_70" alt="在这里插入图片描述"><br>代码使用的tensorflow，是1.13.1版本的；需要将cuda降到10.0。<br>这里有一篇<a href="https://blog.csdn.net/omodao1/article/details/83241074">博客</a>，展示了Tensorflow不同版本要求与CUDA及CUDNN版本对应关系，一般我们都是装好了tensorflow-gpu，然后需要对应版本的CUDA及CUDNN，可以让大家参考一下。</p>
<p>用conda安装对应的cudatoolkit</p>
<figure class="highlight bash"><table><tr><td class="code"><pre><span class="line">conda install cudatoolkit=10.0 -c https://mirrors.tuna.tsinghua.edu.cn/anaconda/pkgs/free/linux-64/</span><br></pre></td></tr></table></figure>
<p>用conda安装对应的cudnn</p>
<figure class="highlight bash"><table><tr><td class="code"><pre><span class="line">conda install cudnn=7.5.0 -c https://mirrors.tuna.tsinghua.edu.cn/anaconda/pkgs/main/linux-64/</span><br></pre></td></tr></table></figure>
<p>这样每次运行时，会链接到这个版本的cuda和cudnn，实际的cuda版本没有变。</p>
<p>参考链接：<br><a href="https://www.cnblogs.com/sohuhome/p/11704529.html">tf-gpu报错：ImportError: libcublas.so.10.0: cannot open shared object file: No such file or directory</a><br><a href="https://blog.csdn.net/weixin_40588315/article/details/85881338?utm_medium=distribute.pc_relevant.none-task-blog-BlogCommendFromMachineLearnPai2-1.nonecase&depth_1-utm_source=distribute.pc_relevant.none-task-blog-BlogCommendFromMachineLearnPai2-1.nonecase">Ubuntu下使用conda在虚拟环境中安装CUDA、CUDNN及Tensorflow</a><br><a href="https://blog.csdn.net/chr1341901410/article/details/104903016?utm_medium=distribute.pc_relevant.none-task-blog-BlogCommendFromMachineLearnPai2-1.nonecase&depth_1-utm_source=distribute.pc_relevant.none-task-blog-BlogCommendFromMachineLearnPai2-1.nonecase">conda安装cudnn和cudatoolkit</a></p>
]]></content>
</entry>
<entry>
<title>如何批量删除PDF文件中所有注释/高亮</title>
<url>/2022/01/04/%E5%A6%82%E4%BD%95%E6%89%B9%E9%87%8F%E5%88%A0%E9%99%A4PDF%E6%96%87%E4%BB%B6%E4%B8%AD%E6%89%80%E6%9C%89%E6%B3%A8%E9%87%8A:%E9%AB%98%E4%BA%AE/</url>
<content><![CDATA[<h1 id="摘要"><a href="#摘要" class="headerlink" title="摘要"></a>摘要</h1><p>有时候我们拿到一份PDF学习资料，发现上面有前人留下的很多注释/高亮，干扰了我们对资料的阅读。对于一条注释/高亮，我们可以选择直接删除；对于多条注释/高亮，我们可以选择全选删除。遗憾的是，直接使用ctrl+A无法全选所有注释/高亮。这里介绍一种方法，使用shift键，只需4步批量删除PDF文件中所有注释/高亮。</p>
<span id="more"></span>
<h2 id="方法"><a href="#方法" class="headerlink" title="方法"></a>方法</h2><ol>
<li>使用Adobe Acrobat打开PDF文件，点击右侧黄色的“注释”功能；</li>
<li>按照时间或类型排序，单击鼠标左键选择需要批量删除的注释/高亮的第一条；</li>
<li>按住shift键，单击鼠标左键选择需要批量删除的注释/高亮的最后一条；</li>
<li>按删除键删除，鼠标右键选择“删除”。</li>
</ol>
<p>以上是从Adobe Support Community找到的答案，在这里给出社区链接<a href="#refer-anchor-1"><sup>1</sup></a><sup>,</sup><a href="#refer-anchor-2"><sup>2</sup></a>作为参考。</p>
<h2 id="参考"><a href="#参考" class="headerlink" title="参考"></a>参考</h2><div id="refer-anchor-1"></div>
[1] [How to delete ALL highlighted text in pdf](https://community.adobe.com/t5/acrobat-reader-discussions/how-to-delete-all-highlighted-text-in-pdf/m-p/8361894)
<div id="refer-anchor-2"></div>
[2] [Can I delete all comments on a PDF with one click?](https://community.adobe.com/t5/acrobat-discussions/can-i-delete-all-comments-on-a-pdf-with-one-click/m-p/10472171)]]></content>
<categories>
<category>计算机技术</category>
</categories>
<tags>
<tag>PDF</tag>
</tags>
</entry>
</search>