java - Checking Palindromes using BigInteger -

i working on /r/dailyprogramming challenge calculate how many steps takes turn number palindromic number. e.g. 24 gets palindromic after 1 steps: 66 -> 24 + 42 = 66.

the program working however, 1 of inputs quite large, have chosen use biginteger solve overflow problem having regular integers.

originally using code similar found on leetcode check if value palindrome, worked input didn't cause overflow:

original code example used check if palindrome:

public int reverse(int x) {      //flag marks if x negative     boolean flag = false;      if (x < 0) {         x = 0 - x;         flag = true;     }      int res = 0;     int p = x;      while (p > 0) {         int mod = p % 10;         p = p / 10;         res = res * 10 + mod;     }      if (flag) {         res = 0 - res;     }      return res; } 

i have since updated code account use of biginteger's. updated code loops indefinitely, palindrome never detected, continues reverse , add - have stepped through program few times , unable determine have incorrectly transposed original code have written.

making_numbers_palindromic:

public class making_numbers_palindromic {  private biginteger originalnumber; private int count = 0;  public void run(biginteger number) {     assert (checkpositive(number));      originalnumber = number;      boolean flag = false;     while (!flag) {         if (checkpalindrome(number)) {             system.out.printf("%d gets palindromic after" +                    "%d steps: %d", originalnumber, count, number);             system.exit(1);         }         number = addreverse(number);         count++;     } }  /**  * @param number  * @return biginteger#signum returns 1 if value positive  */ public boolean checkpositive(biginteger number) {     if (number.compareto(biginteger.zero) == 1) {         return true;     }     return false; }  public boolean checkpalindrome(biginteger number) {     if (number.compareto(biginteger.zero) < 0) return false;      biginteger div = new biginteger("1");      while ((number.divide(div)).compareto(biginteger.ten) >= 0) {         div = div.multiply(biginteger.ten);     }      while (!number.equals(new biginteger("0"))) {         biginteger len = number.divide(div);         biginteger revs = number.mod(biginteger.ten);          if (!len.equals(revs)) return false;          number = (number.mod(div)).divide(biginteger.ten);         div = div.divide(biginteger.ten);     }     return true; }  public biginteger addreverse(biginteger number) {     return number.add(reverse(number)); }  public biginteger reverse(biginteger number) {     boolean flag = false;     biginteger reverse = new biginteger("0");     biginteger numcopy = number;      if (number.compareto(biginteger.zero) == -1) {         flag = true;     }      while (numcopy.compareto(biginteger.zero) > 0) {         biginteger mod = numcopy.mod(biginteger.ten);         numcopy = numcopy.divide(biginteger.ten);         reverse = (reverse.multiply(biginteger.ten)).add(mod);     }      if (flag) {         reverse = biginteger.zero.subtract(reverse);     }     return reverse;   } } 

this first time using biginteger , have taken @ javadocs - cannot see apparent may wrong usage of biginteger#compare or biginteger#equals

the main method:

public class bootstrap {     public static void main(string[] args) {      making_numbers_palindromic mnp = new making_numbers_palindromic();      if (args.length > 0) {          mnp.run(biginteger.valueof(integer.parseint(args[0])));     } else {         system.out.printf("usage: palindromic number");     }   } } 

i plan make making_numbers_palindromic private constructor can use utility class, have not got around yet.

thank-you

not sure if that's error, in checkpalindrome

div = div.divide(biginteger.ten); 

should be

div = div.divide(biginteger.valueof(100)); 

since after compare first , last digits of number, discard them, div of next iterations should 100 times smaller current div.

that said, after fix bugs, doing double work should doing, since if reverse number, can use reversed number determine if original palindrome (by comparing original reversed number).