README.md

comments	difficulty	edit_url	tags
true	中等	https://github.com/doocs/leetcode/edit/main/solution/0100-0199/0186.Reverse%20Words%20in%20a%20String%20II/README.md	双指针 字符串

186. 反转字符串中的单词 II 🔒

English Version

题目描述

给你一个字符数组 s ，反转其中 单词 的顺序。

单词 的定义为：单词是一个由非空格字符组成的序列。s 中的单词将会由单个空格分隔。

必须设计并实现 原地 解法来解决此问题，即不分配额外的空间。

 

示例 1：

输入：s = ["t","h","e"," ","s","k","y"," ","i","s"," ","b","l","u","e"]
输出：["b","l","u","e"," ","i","s"," ","s","k","y"," ","t","h","e"]

示例 2：

输入：s = ["a"]
输出：["a"]

 

提示：

• 1 <= s.length <= 105
• s[i] 可以是一个英文字母（大写或小写）、数字、或是空格 ' ' 。
• s 中至少存在一个单词
• s 不含前导或尾随空格
• 题目数据保证：s 中的每个单词都由单个空格分隔

解法

方法一：双指针

我们可以遍历字符数组 $s$，利用双指针 $i$ 和 $j$ 找到每个单词的起始位置和结束位置，然后反转每个单词，最后再反转整个字符数组。

时间复杂度 $O(n)$，其中 $n$ 为字符数组 $s$ 的长度。空间复杂度 $O(1)$。

Python3

class Solution:
    def reverseWords(self, s: List[str]) -> None:
        def reverse(i: int, j: int):
            while i < j:
                s[i], s[j] = s[j], s[i]
                i, j = i + 1, j - 1

        i, n = 0, len(s)
        for j, c in enumerate(s):
            if c == " ":
                reverse(i, j - 1)
                i = j + 1
            elif j == n - 1:
                reverse(i, j)
        reverse(0, n - 1)

Java

class Solution {
    public void reverseWords(char[] s) {
        int n = s.length;
        for (int i = 0, j = 0; j < n; ++j) {
            if (s[j] == ' ') {
                reverse(s, i, j - 1);
                i = j + 1;
            } else if (j == n - 1) {
                reverse(s, i, j);
            }
        }
        reverse(s, 0, n - 1);
    }

    private void reverse(char[] s, int i, int j) {
        for (; i < j; ++i, --j) {
            char t = s[i];
            s[i] = s[j];
            s[j] = t;
        }
    }
}

C++

class Solution {
public:
    void reverseWords(vector<char>& s) {
        auto reverse = [&](int i, int j) {
            for (; i < j; ++i, --j) {
                swap(s[i], s[j]);
            }
        };
        int n = s.size();
        for (int i = 0, j = 0; j < n; ++j) {
            if (s[j] == ' ') {
                reverse(i, j - 1);
                i = j + 1;
            } else if (j == n - 1) {
                reverse(i, j);
            }
        }
        reverse(0, n - 1);
    }
};

Go

func reverseWords(s []byte) {
	reverse := func(i, j int) {
		for ; i < j; i, j = i+1, j-1 {
			s[i], s[j] = s[j], s[i]
		}
	}
	i, n := 0, len(s)
	for j, c := range s {
		if c == ' ' {
			reverse(i, j-1)
			i = j + 1
		} else if j == n-1 {
			reverse(i, j)
		}
	}
	reverse(0, n-1)
}

TypeScript

/**
 Do not return anything, modify s in-place instead.
 */
function reverseWords(s: string[]): void {
    const n = s.length;
    const reverse = (i: number, j: number): void => {
        for (; i < j; ++i, --j) {
            [s[i], s[j]] = [s[j], s[i]];
        }
    };
    for (let i = 0, j = 0; j <= n; ++j) {
        if (s[j] === ' ') {
            reverse(i, j - 1);
            i = j + 1;
        } else if (j === n - 1) {
            reverse(i, j);
        }
    }
    reverse(0, n - 1);
}