69. Sqrt(x) (Divide-and-Conquer)

Implement int sqrt(int x).

Compute and return the square root of x.

注意: 计算平方的时候可能会溢出，所以mid要定义为long

另外，二分法初始上限不可能超过n/2+1

class Solution {
public:
    int mySqrt(int x) {
 
        int left = ;
        int right = x/+;
 
        while (left <= right)
        {
            long mid = left + (right - left) / ;
            if (mid * mid <= x && (mid + ) * (mid + ) > x)
            {
                return mid;
            }
            else if (mid * mid < x)
                left = mid + ;
            else
                right = mid - ;
        }
 
        return -;
    }
};