leetcode-x的平方根

题目

https://leetcode-cn.com/problems/sqrtx/

思路

使i从1开始查找，判断i*i > x时，循环结束，当i*i<=x，记录res的值，这整个过程，可以用二分查找的思想我直接暴力😭😭，思维还不够敏捷。

AC代码

class Solution {
public:
    int mySqrt(int x) {
        /*
        long long  res;
        if (x == 0) return 0; 
        if(x == 1) return 1;
        long long i = 1;
        while (i*i <= x){
            res = i;
            i++; 
        }
        return res;
        */
        //二分法
        int left = 0;
        int right = x;
        int res;
        if (x == 0) return 0;
        if (x == 1) return 1;
        while (left <= right){
            int mid = (right - left) / 2 + left; //防止溢出
            if ((long long) mid*mid <= x){
                res = mid; 
                left = mid+1;
            }
            else
                right = mid-1;    
        }
        return res; //对从1~n遍历查找想要的值，应该使用二分查找

    }
};