Java Algorithms How to - Compute the Palindrome of a number by adding the number composed of

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Question

We would like to know how to compute the Palindrome of a number by adding the number composed of.

Answer

    /*from   w w  w  . jav  a 2 s  .co  m*/
import java.math.BigInteger;

/** Compute the Palindrome of a number by adding the number composed of
 * its digits in reverse order, until a Palindrome occurs.
 * e.g., 42->66 (42+24); 1951->5995 (1951+1591=3542; 3542+2453=5995).
 * <P>TODO: Do we need to handle negative numbers?
 * @author Ian Darwin, http://www.darwinsys.com/
 * @version $Id: PalindromeBig.java,v 1.3 2004/02/09 03:33:57 ian Exp $.
 */
public class Main {

  public static boolean verbose = true;

  public static void main(String[] argv) {
       BigInteger l = new BigInteger("9999999");
     System.out.println(findPalindrome(l));
    
  }

  /** find a palindromic number given a starting point, by
   * calling ourself until we get a number that is palindromic.
   */
  public static BigInteger findPalindrome(BigInteger num) {
    if (num.compareTo(BigInteger.ZERO) < 0)
      throw new IllegalStateException("negative");
    if (isPalindrome(num))
      return num;
    if (verbose)
      System.out.println("Trying " + num);
    return findPalindrome(num.add(reverseNumber(num)));
  }

  /** A ridiculously large number  */
  protected static final int MAX_DIGITS = 255;

  /** Check if a number is palindromic. */
  public static boolean isPalindrome(BigInteger num) {
    String digits = num.toString();
    int numDigits = digits.length();
    if (numDigits >= MAX_DIGITS) {
      throw new IllegalStateException("too big");
    }
    // Consider any single digit to be as palindromic as can be
    if (numDigits == 1)
      return true;
    for (int i=0; i<numDigits/2; i++) {
      // System.out.println(
      //   digits.charAt(i) + " ? " + digits.charAt(numDigits - i - 1));
      if (digits.charAt(i) != digits.charAt(numDigits - i - 1))
        return false;
    }
    return true;
  }

  static BigInteger reverseNumber(BigInteger num) {
    String digits = num.toString();
      int numDigits = digits.length();
    char[] sb = new char[numDigits];
    for (int i=0; i<digits.length(); i++) {
      sb[i] = digits.charAt(numDigits - i - 1);
    }
    // Debug.println("rev",
    //   "reverseNumber(" + digits + ") -> " + "\"" + sb + "\"");
    return new BigInteger(new String(sb));
  }
}

The code above generates the following result.

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