leetcode Sqrt(x) Binary Search

Pay attention to that mid * mid may be larger than int. So long long is enough. 

class Solution { public:  int sqrt(int x) {    // Note: The Solution object is instantiated only once and is reused by each test case.    int l = 0, r = x;    while (l <= r) {      int mid = (l + r) / 2;      long long mid2 = (long long)mid * (long long)mid;      long long mid12 = (long long)(mid + 1) * (long long)(mid + 1);            if (mid2 <= x && mid12 > x)        return (int)mid;      if (mid2 > x)        r = mid -1;      else        l = mid +1;    }  }};