5 COMMENTS

1. 

   Ash Winchester Posted on July 11, 2020 at 6:49 pm

   bad solution, do not use this. instead check this solution:

   class Solution {
   public int mySqrt(int x) {
   if (x < 2) return x;

   long num;
   int pivot, left = 2, right = x / 2;
   while (left <= right) {
   pivot = left + (right – left) / 2;
   num = (long)pivot * pivot;
   if (num > x) right = pivot – 1;
   else if (num < x) left = pivot + 1;
   else return pivot;
   }

   return right;
   }
   }

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2. 

   Mr. Mystiks Posted on July 11, 2020 at 6:49 pm

   This code actually doesn’t work for the test case: X=8. It goes into an infinite loop. This is garbage code and I don’t think you even understand the solution.

   Reply
3. 

   Ashwini Khedkar Posted on July 11, 2020 at 6:49 pm

   For the case when x = 8, above code, goes into an infinite loop causing time limit to exceed.

   Reply
4. 

   Abdelrahman Posted on July 11, 2020 at 6:49 pm

   why did you use lo+1 ?

   Reply
5. 

   Kun Jiang Posted on July 11, 2020 at 6:49 pm

   看了很多不同的视频 你的概念思路是讲的最好的 非常清楚 感谢 请继续 并且可以把70题加上吗？

   Reply