Leetcode: Valid Perfect Square

Given a positive integer num, write a function which returns True if num is a perfect square else False.

Note: Do not use any built-in library function such as sqrt.

Example 1:

Input: 16
Returns: True
Example 2:

Input: 14
Returns: False

我的binary Search 解法：无需变成long

public class Solution {
    public boolean isPerfectSquare(int num) {
        if (num <= 0) return false;
        int l=1, r=num/2+1;
        while (l <= r) {
            int mid = l + (r-l)/2;
            if (mid == num/mid && num%mid == 0) return true;
            else if (mid > num/mid) {
                r = mid - 1;
            }
            else l = mid + 1;
        }
        return false;
    }
}

别人三种方法总结：

1. a square number is 1+3+5+7+... Time Complexity O(sqrt(N)) (Credit to lizhibupt, thanks for correcting this).
2. binary search. Time Complexity O(logN)
3. Newton Method. See [this wiki page][1]. Time Complexity is close to constant, given a positive integer.

public boolean isPerfectSquare(int num) {
      if (num < 1) return false;
      for (int i = 1; num > 0; i += 2)
        num -= i;
      return num == 0;
    }
    
    public boolean isPerfectSquare(int num) {
      if (num < 1) return false;
      long left = 1, right = num;// long type to avoid 2147483647 case
    
      while (left <= right) {
        long mid = left + (right - left) / 2;
        long t = mid * mid;
        if (t > num) {
          right = mid - 1;
        } else if (t < num) {
          left = mid + 1;
        } else {
          return true;
        }
      }
    
      return false;
    }
    
    boolean isPerfectSquare(int num) {
      if (num < 1) return false;
      long t = num / 2 + 1; //or t = num as the begining
      while (t * t > num) {
        t = (t + num / t) / 2;
      }
      return t * t == num;
    }