69. Sqrt(x)

Implement int sqrt(int x).

Compute and return the square root of x.

思路: sqrt（x）肯定是小于x的大于0的数，所以可以用二分查找。注意最后返回high的值是说如果x没有正数的平方根取最接近于小于mid的那个数。注意mid要用long，防止mid*mid越界。
http://www.cnblogs.com/springfor/p/3857772.html

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