DevExplained

9. Palindrome Number

Math

Explanation:

To determine if a given integer is a palindrome without converting it to a string, we can reverse half of the number and compare it with the other half.

  1. For negative numbers or numbers ending in 0 (except 0 itself), they cannot be palindromes.
  2. Initialize a variable reversed to store the reversed half of the number.
  3. While x is greater than reversed, reverse the last digit of x and add it to reversed.
  4. If x has an odd number of digits, we can simply check if x == reversed/10.
  5. If x has an even number of digits, we can check if x == reversed.

This approach avoids the need to convert the integer to a string.

Time complexity: O(log(x)) where x is the input number. Space complexity: O(1)

:

View Leetcode Problem
class Solution {
    public boolean isPalindrome(int x) {
        if (x < 0 || (x % 10 == 0 && x != 0)) {
            return false;
        }
        
        int reversed = 0;
        while (x > reversed) {
            int lastDigit = x % 10;
            reversed = reversed * 10 + lastDigit;
            x /= 10;
        }
        
        return x == reversed || x == reversed / 10;
    }
}

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