Palindrome Number

9.Palindrome Number

Math

Two Pointers

Problem Statement:

Given an integer x, return true if x is a palindrome, and false otherwise. An integer is a palindrome when it reads the same forward and backward.

Example:

Input:

x = 121

→

Output:

true

121 reads as 121 from left to right and right to left.

Algorithm:

1. Check if number is negative (not palindrome)
2. Store original number for comparison
3. Build reverse of number by extracting digits
4. Compare original with reversed number

Complexity:

Time: O(log n) | Space: O(1)

Java Solution:

public boolean isPalindrome(int x) {
    int num = x;
    int n = 0;
    
    // Negative numbers are not palindromes
    if (x < 0) return false;
    
    while (x != 0) {
        // Build reversed number digit by digit
        int last = x % 10;
        n = n * 10 + last;
        x /= 10;
    }
    
    return n == num;
}

Python Solution:

def isPalindrome(self, x: int) -> bool:
    # Negative numbers are not palindromes
    if x < 0:
        return False
        
    num, n = x, 0
    
    while x:
        # Build reversed number digit by digit
        n = n * 10 + x % 10
        x //= 10
        
    return n == num

C++ Solution:

bool isPalindrome(int x) {
    // Negative numbers are not palindromes
    if (x < 0) return false;
    
    long num = x;
    long n = 0;
    
    while (x != 0) {
        // Build reversed number digit by digit
        n = n * 10 + x % 10;
        x /= 10;
    }
    
    return n == num;
}