Binary Solution - Sqrt(x) - Iteration Binary

The idea is to binary try square and adjust high or low based on comparison. Instead of using recursive, iteration is used here for binary search: if low + 1 < high, high=mid/low=mid/return.

Notice that the while condition mush be high - low > 1, as the case 2 will cause loop forever if using h>l only!

public class Solution {
    public int sqrt(int x) {
        if(x <= 0){
            return 0;
        }
        if(x == 1){
            return 1;
        }
        long low = 1;
        long high = x;
        long mid = (low + high) / 2;
       
        while(high - low > 1){
            mid = (low + high) / 2;
            if(mid * mid > x){
                high = mid;
            }else if(mid * mid < x){
                low = mid;
            }else{
                return (int)mid;
            }
        }
        return (int)low;
    }
}