How do I check if a number is a palindrome?

Question

How do I check if a number is a palindrome?

Any language. Any algorithm. (except the algorithm of making the number a string and then reversing the string).

Answer 1

Here is an Scheme version that constructs a function that will work against any base. It has a redundancy check: return false quickly if the number is a multiple of the base (ends in 0). And it doesn't rebuild the entire reversed number, only half. That's all we need.

(define make-palindrome-tester
   (lambda (base)
     (lambda (n)
       (cond
         ((= 0 (modulo n base)) #f)
         (else
          (letrec
              ((Q (lambda (h t)
                    (cond
                      ((< h t) #f)
                      ((= h t) #t)
                      (else
                       (let* 
                           ((h2 (quotient h base))
                            (m  (- h (* h2 base))))
                         (cond 
                           ((= h2 t) #t)
                           (else
                            (Q h2 (+ (* base t) m))))))))))           
            (Q n 0)))))))

Answer 2

Above most of the answers having a trivial problem is that the int variable possibly might overflow.

Refer to http://leetcode.com/2012/01/palindrome-number.html

boolean isPalindrome(int x) {
    if (x < 0)
        return false;
    int div = 1;
    while (x / div >= 10) {
        div *= 10;
    }
    while (x != 0) {
        int l = x / div;
        int r = x % 10;
        if (l != r)
            return false;
        x = (x % div) / 10;
        div /= 100;
    }
    return true;
}

Answer 3

I always use this python solution due to its compactness.

def isPalindrome(number):
    return int(str(number)[::-1])==number

Answer 4

checkPalindrome(int number)
{
    int lsd, msd,len;
    len = log10(number);
    while(number)
    {
        msd = (number/pow(10,len)); // "most significant digit"
        lsd = number%10; // "least significant digit"
        if(lsd==msd)
        {
            number/=10; // change of LSD
            number-=msd*pow(10,--len); // change of MSD, due to change of MSD
            len-=1; // due to change in LSD
            } else {return 1;}
    }
    return 0;
}

Answer 5

For any given num:

 n = num;
 rev = 0;
 while (num > 0)
 {
      dig = num % 10;
      rev = rev * 10 + dig;
      num = num / 10;
 }

If n == rev then num is a palindrome:

cout << "Number " << (n == rev ? "IS" : "IS NOT") << " a palindrome" << endl;

Answer 6

Golang version:

package main

import "fmt"

func main() {
    n := 123454321
    r := reverse(n)
    fmt.Println(r == n)
}

func reverse(n int) int {
    r := 0
    for {
        if n > 0 {
            r = r*10 + n%10         
            n = n / 10
        } else {
            break
        }
    }
    return r
}