假设你正在读取一串整数。每隔一段时间，你希望能找出数字 x 的秩(小于或等于 x 的值的个数)。请实现数据结构和算法来支持这些操作，也就是说：
实现 track(int x) 方法，每读入一个数字都会调用该方法；
实现 getRankOfNumber(int x) 方法，返回小于或等于 x 的值的个数。
注意：本题相对原题稍作改动
示例:
输入: ["StreamRank", "getRankOfNumber", "track", "getRankOfNumber"] [[], [1], [0], [0]] 输出: [null,0,null,1]
提示：
x <= 50000track和getRankOfNumber方法的调用次数均不超过 2000 次
方法一：树状数组
树状数组，也称作“二叉索引树”（Binary Indexed Tree）或 Fenwick 树。 它可以高效地实现如下两个操作：
- 单点更新
update(x, delta)： 把序列 x 位置的数加上一个值 delta； - 前缀和查询
query(x)：查询序列[1,...x]区间的区间和，即位置 x 的前缀和。
这两个操作的时间复杂度均为
树状数组最基本的功能就是求比某点 x 小的点的个数（这里的比较是抽象的概念，可以是数的大小、坐标的大小、质量的大小等等）。
比如给定数组 a[5] = {2, 5, 3, 4, 1}，求 b[i] = 位置 i 左边小于等于 a[i] 的数的个数。对于此例，b[5] = {0, 1, 1, 2, 0}。
解决方案是直接遍历数组，每个位置先求出 query(a[i])，然后再修改树状数组 update(a[i], 1) 即可。当数的范围比较大时，需要进行离散化，即先进行去重并排序，然后对每个数字进行编号。
class BinaryIndexedTree:
def __init__(self, n):
self.n = n
self.c = [0] * (n + 1)
@staticmethod
def lowbit(x):
return x & -x
def update(self, x, delta):
while x <= self.n:
self.c[x] += delta
x += BinaryIndexedTree.lowbit(x)
def query(self, x):
s = 0
while x > 0:
s += self.c[x]
x -= BinaryIndexedTree.lowbit(x)
return s
class StreamRank:
def __init__(self):
self.tree = BinaryIndexedTree(50010)
def track(self, x: int) -> None:
self.tree.update(x + 1, 1)
def getRankOfNumber(self, x: int) -> int:
return self.tree.query(x + 1)
# Your StreamRank object will be instantiated and called as such:
# obj = StreamRank()
# obj.track(x)
# param_2 = obj.getRankOfNumber(x)class BinaryIndexedTree {
private int n;
private int[] c;
public BinaryIndexedTree(int n) {
this.n = n;
c = new int[n + 1];
}
public static int lowbit(int x) {
return x & -x;
}
public void update(int x, int delta) {
while (x <= n) {
c[x] += delta;
x += lowbit(x);
}
}
public int query(int x) {
int s = 0;
while (x > 0) {
s += c[x];
x -= lowbit(x);
}
return s;
}
}
class StreamRank {
private BinaryIndexedTree tree;
public StreamRank() {
tree = new BinaryIndexedTree(50010);
}
public void track(int x) {
tree.update(x + 1, 1);
}
public int getRankOfNumber(int x) {
return tree.query(x + 1);
}
}
/**
* Your StreamRank object will be instantiated and called as such:
* StreamRank obj = new StreamRank();
* obj.track(x);
* int param_2 = obj.getRankOfNumber(x);
*/class BinaryIndexedTree {
public:
int n;
vector<int> c;
BinaryIndexedTree(int _n)
: n(_n)
, c(_n + 1) { }
void update(int x, int delta) {
while (x <= n) {
c[x] += delta;
x += lowbit(x);
}
}
int query(int x) {
int s = 0;
while (x > 0) {
s += c[x];
x -= lowbit(x);
}
return s;
}
int lowbit(int x) {
return x & -x;
}
};
class StreamRank {
public:
BinaryIndexedTree* tree;
StreamRank() {
tree = new BinaryIndexedTree(50010);
}
void track(int x) {
tree->update(x + 1, 1);
}
int getRankOfNumber(int x) {
return tree->query(x + 1);
}
};
/**
* Your StreamRank object will be instantiated and called as such:
* StreamRank* obj = new StreamRank();
* obj->track(x);
* int param_2 = obj->getRankOfNumber(x);
*/type BinaryIndexedTree struct {
n int
c []int
}
func newBinaryIndexedTree(n int) *BinaryIndexedTree {
c := make([]int, n+1)
return &BinaryIndexedTree{n, c}
}
func (this *BinaryIndexedTree) lowbit(x int) int {
return x & -x
}
func (this *BinaryIndexedTree) update(x, delta int) {
for x <= this.n {
this.c[x] += delta
x += this.lowbit(x)
}
}
func (this *BinaryIndexedTree) query(x int) int {
s := 0
for x > 0 {
s += this.c[x]
x -= this.lowbit(x)
}
return s
}
type StreamRank struct {
tree *BinaryIndexedTree
}
func Constructor() StreamRank {
tree := newBinaryIndexedTree(50010)
return StreamRank{tree}
}
func (this *StreamRank) Track(x int) {
this.tree.update(x+1, 1)
}
func (this *StreamRank) GetRankOfNumber(x int) int {
return this.tree.query(x + 1)
}
/**
* Your StreamRank object will be instantiated and called as such:
* obj := Constructor();
* obj.Track(x);
* param_2 := obj.GetRankOfNumber(x);
*/