Sqrt(x)

Implement int sqrt(int x).

Compute and return the square root of x.

好的解法： 二分法，注意边界条件。当不满足循环条件时，low>high，经过验证，low-1为正确答案。时间：16ms

class Solution {
public:
    int mySqrt(int x) {
        if(x == 0) return 0;
        if(x == 1) return 1;
        
        int low = 0, high = x;
        int mid = (low + high) / 2;
        
        while(low <= high){
            if(x / mid == mid) return mid;
            else if(x / mid < mid) high = mid - 1;
            else low = mid + 1;
            
            mid = (low + high) / 2;
        }
        return low - 1;
        
    }
};

自己写的：调用系统函数。面试肯定过不了啊。。时间：14ms。是怎么实现的呢？？

class Solution {
public:
    int mySqrt(int x) {
        return floor(sqrt(x*1.00));
    }
};