LeetCode69 Sqrt(x)

Implement int sqrt(int x).

Compute and return the square root of x, where x is guaranteed to be a non-negative integer.

Since the return type is an integer, the decimal digits are truncated and only the integer part of the result is returned.

Example 1:

Input: 4
Output: 2
Example 2:

Input: 8
Output: 2
Explanation: The square root of 8 is 2.82842…, and since
the decimal part is truncated, 2 is returned.

public int mySqrt(int x) {
        if(x == 0 || x == 1) return x;
        int l = 1;
        int r = x / 2;
        while(l <= r){
            int mid = l + (r - l) / 2;
            if(x / mid == mid) return mid;
            else if(x / mid > mid) l = mid + 1;
            else r = mid - 1;
        }
        return l-1;
    }