Difficulty: Medium

Implement int sqrt(int x) . Compute and return the square root of x.

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public int mySqrt(int x) {
         if (x == 0)
            return 0;
        //Instead of using fancy Newton's method, this plain binary search approach also works.
        // https://discuss.leetcode.com/topic/8680/a-binary-search-solution
        if (x < 0)  throw new IllegalArgumentException ("x is less than 0");
        int start = 1, end = Integer.MAX_VALUE;
        while (true) {
            int mid = start + (end - start) / 2;
            
            if (mid > x / mid) {
              end = mid - 1;
            } else {
                if (mid + 1 > x /(mid + 1)){
                    return mid;
                }
                start = mid + 1;
            }
        }
    }