LeetCode OJ-- Sqrt(x) *

https://oj.leetcode.com/problems/sqrtx/

求一个数的开方，二分查找一个数，直到这个数的平方 - target 小于 0.001。

但这道题出的不好，返回值竟然是 int ,脑洞太大了。

class Solution {
public:
    int sqrt(int x) {
        if(x<=0)
            return 0;
        if(x == 1)
            return 1;
        double fi = 0.0001;
        //binary search
        double low = 0, high = x, middle = x/2;
        while(1)
        {
            if(abs(middle*middle - x) < fi)
                return middle;
            if(middle*middle > x)
                high = middle;
            else if(middle*middle<x)
                low = middle;
            middle= (low + high)/2.0;
        }
    }
};