Find Next Palindrome Number

Given a number, find the next smallest palindrome larger than the number. For example if the number is 125, next smallest palindrome is 131.
Java program : 

public class NextPalindrome {
 public static void main(String[] args) {
  int n = 2133;

  // temp variable to perform operations on n
  int temp = n;
  // calculate the number of digits
  int digits = 0;
  while (temp != 0) {
   temp = temp / 10;
   digits++;
  }

  // edge case if the given number contains all 9 s like 9 or 99 or 999
  // or...,round off to next number
  int divider = (int) Math.pow(10, digits);
  if ((n + 1) % divider == 0) {
   n = n + 1;
   digits++;
  }
  if (digits == 0)
   return;

  // handling the special case of single digit numbers
  if (digits == 1) {
   System.out.println("Next higher palindrome is : " + n + 1);
   return;
  }

  // create array of size digits to store the digits of the given number
  int[] a = new int[digits];
  temp = n;
  for (int i = digits - 1; i >= 0 && temp != 0; i--) {
   a[i] = temp % 10;
   temp = temp / 10;
  }

  // now array contains the digits in same order as of number
  int nextPalindrome = 0;
  boolean isPalindrome = false;// used for track whether we got palindrome
  // or not
  while (!isPalindrome) {
   // create mirror image of the given number
   createMirrorImage(a);
   nextPalindrome = constructNumber(a);
   if (nextPalindrome <= n) {
    int middle = (int) Math.ceil(digits / 2);
    // if the middle digit is 9 then round off to the next higher
    // number
    if (digits % 2 == 0) {
     if (a[digits / 2] == 9) {
      a[middle] = 0;
      a[middle - 1] = 0;
      a[middle - 2] += 1;
     } else {
      a[middle] += 1;
      a[middle - 1] += 1;
     }
    } else {
     if (a[middle] == 9) {
      a[middle] = 0;
      a[middle - 1] = a[middle - 1] + 1;
     } else {
      a[middle] += 1;
     }

    }
   } else {
    isPalindrome = true;
   }
  }
  System.out.println("Next higher palindrome is : " + nextPalindrome);
 }

 private static int constructNumber(int[] a) {
  int num = 0;
  for (int i = 0; i < a.length; i++) {
   num = num * 10 + a[i];
  }
  return num;
 }

 private static void createMirrorImage(int[] a) {
  int length = a.length;
  for (int i = 0, j = length - 1; i <= j; i++, j--) {
   a[j] = a[i];
  }
 }
}

Refer :  http://www.ardendertat.com/2011/12/01/programming-interview-questions-19-find-next-palindrome-number/