LeetCode之Sqrt(x)

/*通过二分法搜索。但是要注意两个数相乘可能会大于int能表示的最大值，
所以程序中采用long long型数据。*/
class Solution {
public:
    int mySqrt(int x) {
        if(x == 0) return 0;
        long long left = 1, right = x;
        while(left < right){
            long long mid = (left + right) / 2;
            if(mid * mid == x) return mid;
            else if(mid * mid < x){
                if((mid+1)*(mid+1) > x) return mid;
                else left = mid + 1;
            }
            else{
                if((mid-1)*(mid-1) < x) return mid-1;
                else right = mid - 1;
            }
        }
    }
};