LeetCode 69 Sqrt(x)

int left=1;
int right=x;
while(left<right-1){
  int mid=left+(right-left)/2;
  int tmp=mid*mid;   //Might be Overflow   
  if(tmp==x){
    return tmp;  
  }else if(tmp>x){
     right=mid-1;
  }else{
     left=mid;
  }    
}

public int sqrt(int x) {
        // Corner case
        if (x <= 1) {
            return x;
        }
        int left = 1;
        int right = x;
        while (left < right - 1) {
            int mid = left + (right - left) / 2;
            int tmp = x / mid;
            if (tmp == mid) {
                return mid;
            } else if (tmp > mid) {
                left = mid;
            } else {
                right = mid - 1;
            }
        }
        // Post Processing
        if (x / right >= right) {
            return right;
        }
        return left;
    }