Palindrome chain length in Java

The challenge

Number is a palindrome if it is equal to the number with digits in reversed order. For example, 5441714884 are palindromes, and 431944773 are not.

Write a function which takes a positive integer and returns the number of special steps needed to obtain a palindrome. The special step is: “reverse the digits, and add to the original number”. If the resulting number is not a palindrome, repeat the procedure with the sum until the resulting number is a palindrome.

If the input number is already a palindrome, the number of steps is 0.

All inputs are guaranteed to have a final palindrome which does not overflow long.

Example

For example, start with 87:

  87 +   78 =  165     - step 1, not a palindrome
 165 +  561 =  726     - step 2, not a palindrome
 726 +  627 = 1353     - step 3, not a palindrome
1353 + 3531 = 4884     - step 4, palindrome!

4884 is a palindrome and we needed 4 steps to obtain it, so answer for 87 is 4.

Additional info

Some interesting information on the problem can be found in this Wikipedia article on Lychrel numbers.

The solution in Java code

Option 1:

public class Palindromes {
  
  public static int palindromeChainLength (long n) {      
    String ns = "" + n, nrs = "" + new StringBuilder(ns).reverse();
    return ns.equals(nrs) ? 0 : 1 + palindromeChainLength(n + Long.valueOf(nrs));
  }
  
}

Option 2:

public class Palindromes {
    public static int palindromeChainLength(long n) {
        int step = 0;
        while (n != reverseNumber(n)) {
            n = n + reverseNumber(n);
            step++;
        }
        return step;
    }

    private static long reverseNumber(long n) {
        char[] number = String.valueOf(n).toCharArray();
        String reverseNumber = "";
        for (int count = number.length - 1; count >= 0; count--) {
            reverseNumber = reverseNumber + number[count];
        }
        return Long.parseLong(reverseNumber);
    }
}

Option 3:

public class Palindromes {
    public static int palindromeChainLength (long n) {
      String s = String.valueOf(n);
      String r = new StringBuilder(String.valueOf(n)).reverse().toString();
      if (s.equals(r)) return 0;
      return 1 + palindromeChainLength(n + Long.parseLong(r));
    }
}

Test cases to validate our solution

import org.junit.Test;
import static org.junit.Assert.*;

public class PalindromesTest {
    
    @Test
    public void testPalindrome() {
        assertEquals(0, Palindromes.palindromeChainLength(1));
        assertEquals(0, Palindromes.palindromeChainLength(88));
        assertEquals(0, Palindromes.palindromeChainLength(393));
    }
    
    @Test
    public void testNonPalindrome() {
        assertEquals(1, Palindromes.palindromeChainLength(10));
        assertEquals(1, Palindromes.palindromeChainLength(134));
        assertEquals(4, Palindromes.palindromeChainLength(87));
        assertEquals(7, Palindromes.palindromeChainLength(2897));
        assertEquals(24, Palindromes.palindromeChainLength(89));
    }
    
}
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