Palindrome Number

easy math palindrome

Problem

Given an integer x, return true if x is palindrome integer. An integer is a palindrome when it reads the same backward as forward. For example, 121 is palindrome while 123 is not.

Inputx = 1221

Outputtrue

Pop the last digit of x and push it onto a reversed half. Stop when x ≤ rev. Then rev equals x or rev/10 equals x (odd-length case).

def is_palindrome(x):
    if x < 0 or (x % 10 == 0 and x != 0):
        return False
    rev = 0
    while x > rev:
        rev = rev * 10 + x % 10
        x //= 10
    return x == rev or x == rev // 10

function isPalindrome(x) {
  if (x < 0 || (x % 10 === 0 && x !== 0)) return false;
  let rev = 0;
  while (x > rev) {
    rev = rev * 10 + (x % 10);
    x = Math.floor(x / 10);
  }
  return x === rev || x === Math.floor(rev / 10);
}

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

bool isPalindrome(int x) {
    if (x < 0 || (x % 10 == 0 && x != 0)) return false;
    int rev = 0;
    while (x > rev) {
        rev = rev * 10 + x % 10;
        x /= 10;
    }
    return x == rev || x == rev / 10;
}

Explanation

A number is a palindrome if it reads the same forwards and backwards. Instead of turning it into a string, we cleverly build up the reverse of only the lower half of the digits and compare it with the upper half.

First, two quick rejects: any negative number is not a palindrome (the minus sign breaks symmetry), and any number ending in 0 (other than 0 itself) cannot be one, because its reverse would start with a leading zero.

Then we loop: each step pops the last digit of x with x % 10 and pushes it onto rev using rev = rev * 10 + x % 10, while shrinking x with x //= 10. We stop once x <= rev, meaning we have processed half the digits.

At the end the number is a palindrome if x == rev (even number of digits) or x == rev // 10 (odd number of digits, where the middle digit sits in rev and is dropped).

Example: x = 1221. We pop 1 then 2: rev becomes 12 and x becomes 12. Now x == rev, so it is a palindrome.

Time: O(log₁₀ x) Space: O(1)