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

 1 class Solution {
 2 public:
 3     int mySqrt(int x) {
 4         if(x == 0) return 0;
 5         if(x == 1) return 1;
 6         
 7         int low = 0, high = x;
 8         int mid = (low + high) / 2;
 9         
10         while(low <= high){
11             if(x / mid == mid) return mid;
12             else if(x / mid < mid) high = mid - 1;
13             else low = mid + 1;
14             
15             mid = (low + high) / 2;
16         }
17         return low - 1;
18         
19     }
20 };

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？？

1 class Solution {
2 public:
3     int mySqrt(int x) {
4         return floor(sqrt(x*1.00));
5     }
6 };