Implementation and analysis of lower_bound and upper_bound

Analyze the comments in the corresponding code.
If x cannot be found, the return value of lower_bound(ll x) is the same as the return value of upper_bound(ll x), which is the first subscript of a number greater than x from left to right

Idea
① Judging the end condition of the loop
② Judging whether left and right should be shifted left and right
③ When determining the position of x after the last dichotomy (x exists)

//设查找的数为x
//x 属于 [lower,upper) 
//二分：每次砍掉一半无用区间，但不砍掉可能存在的答案,  可以自己动手演算一遍，体会过程

int lower(ll x){
    
    //左闭  找区间最左x的数组下标
    int left = 0,right = n-1;//数组从下标0开始储存
    int middle = (left + right)>>1;
    
    while(left <= right){
    
         // ①left == right 时才能取到right 所以要‘=’
                              //由①：②这样最后一次二分开始前一定是left == right == middle(这个条件可以推出下面)
        if(num[middle]<x)    //由①：对任何可能的数据都要使left > right（循环退出) 所以要left right 值都要偏移 
            left = middle+1;  //由②：(如果有x) 不断靠近区间最左x 最后一次二分后x的下标恰好为middle + 1
        else right = middle-1;//由②：(如果有x,求最左所以>=x都要左移) 最后一次二分后num[middle] == x；left = middle 
        middle = (left + right)>>1;
    }
    return left ;//如果有X一定在left
                
}

The code idea is the same as the previous one

upper_bound(ll x){
    
    //右开 求从左到右大于x的第一个数
    int left = 0,right = n - 1;//数组下标n*n不存数
    int middle = (left + right)>>1;
    
    while(left <= right){
    
    //最后一次二分开始前 left == right == middle 
        if(num[middle]<=x)
            left = middle + 1;//右移 
        else right = middle-1;//左移 
        middle = (left + right)>>1;
    }
    return left;//如果有X一定在left
}