leetcode：Sqrt(x) 【Java】

int sqrt(int x)

public class Solution {
    public int mySqrt(int x) {
        if(x < 2){ 
            return x;
        }
        
        int left = 1;
        int right = x / 2;
        int middle = -1;
        int last_middle = -1;
        while(left <= right){
            middle = left + (right - left) / 2;
            if(x / middle > middle){
                left = middle + 1;
                last_middle = middle; //最接近的平方根
            }else if(x / middle < middle){
                right = middle - 1;
            }else{
                return middle;
            }
        }
        return last_middle;
    }
}