The Next Palindrome .. Strange Character Output

Question

I made this code that is supposed to increment a number till get the next palindrome number of of this inputted number.

The program take the number as string " cause it may be a very big digits number ( 0 < digits < 1000000 ) " ....

The Code

int main ()
{
string number = "1243";
int position = number.length()-1;
do
{
    if (number[position] == '9')
    {
        //cout << "hereee";
        number[position] = '0';
        int n1 = (int)number[position-1] - '0';
        n1++;
        number[position-1] = n1 + '0';
        nextPalindrome[position-1];
         cout << number <<"hereee2"<< endl; // only to determine if i get in "if"
    }
    else
    {
        int n1 = (int)number[position] - '0';
        n1++;
        number[position] = n1 + '0';
        cout << number <<"hereee1" << endl; // only to determine if i get in "else"
    }
} while (isPalindrome(number) == false);
}

It's start to take the digit in the current position and increment it and return it as character again

The Problem

cout << number <<"hereee1" << endl;

this line show the number status while runing and it is like that :

12"6 hereee1

12"7 hereee1

12"8 hereee1

12"9 hereee1

12#0 hereee2

12#1 hereee1

while it must to be

1236 hereee1

1237 hereee1

1238 hereee1

1239 hereee1

1240 hereee2

1241 hereee1

i don't know where is the error .. can any one help

NOTE : "isPalindrome" is a function take a string as parameter and if the original string equal to its reverse, it returns true .. else return false

I think there are nicer approaches to this than brute force... — ppeterka
I know there is a lot of ways.. but this a way that i involved in i must find the error reason — Mahmoud
you do realise that doing it this way is 1, unelegant 2, inefficient 3, easy to do wrong and 4, creates problems that wouldn't exist if you did it with being smart? — ppeterka
I am so beginner you know so told me adiiferent way to make it and thanks for advance ^_^ — Mahmoud
You could go with a logic that starts from the middle and goes to the outer edges. If the digit on the right side is greater than the one on the left, add 1 to the left. Then put the same digit to the right position as it is on the left. Repeat until done. Only one special case needs to be handled: if the starting number is full of only 9s, the next palindrome would be in the form 10<.lot of 0s.>01. If I have some time, I think I'll write it into the answer for my own pleasure - it is a good brain exercise. — ppeterka

ppeterka ppeterka · Accepted Answer · 2013-01-29T13:00:23

You don't handle carryover well... This part is invalid:

    int n1 = (int)number[position-1] - '0';
    n1++;
    number[position-1] = n1 + '0';

This just increases the digit on the previous place. And if it happens to be '9' (same as what Mats Petersson tried to suggest), it just overflows... It should however be carried over to the next digit too... This is a recursive solution to this (beware, there might be syntax errors, I haven't coded in C++ since ages...):

/*
* This function adds one to the specified digit of a 
* string containing a decimal integer.
*
* Contains no checks whatsoever. Behavior is undefined when 
* not supplied a valid input string.
*/
int addOneToDigit (string number, int digit)
{
    if (number[digit] == '9')
    {
        number[digit] = '0';
        //we need to handle getting a longer string too...
        if(digit>0) 
        {
            return addOneToDigit(number, digit-1);
        }
        else
        {
            return "1" + number;
        }
    }
    else
    {
        int n1 = (int)number[digit] - '0';
        n1++;
        number[digit] = n1 + '0';
    }
    return number;
}

and the main() would look like something like this:

int main ()
{
    string number = "1243";
    do
    {
      number = addOneToDigit(number,number.length()-1)
    } 
    while (isPalindrome(number) == false);
}

The Next Palindrome .. Strange Character Output

2 Answers