LeetCode OJ：Sqrt(x)（平方根）

Implement int sqrt(int x).

Compute and return the square root of x.

简单的二分法，注意mid应该选为long，否则容易溢出：

class Solution {
public:
    int mySqrt(int x) {
        if(x == 0 || x == 1) return x;
        int beg = 1;
        int end = x;
        long mid = 0;   //这里用long，否则会溢出
        while(beg <= end){
            mid = beg + (end - beg)/2;
            if(mid * mid < x){
                beg = mid + 1;
            }else if(mid * mid > x){
                end = mid - 1;
            }else{
                return mid;
            }
        }
        return end;
    }
};