palindrome

[yeng]

Hi!I'm a sophomore IT student. We are asked to write function that would get as parameter a long integer number and determines whether the number is a palindrome. A number is a palindrome if it reads the same forward and backward. The function Palindrome should be called in the main program. This program somehow involves user define function.
ex: Input(x): 12321, then call PALINDROME(x)
Output : Palindrome : YES

Input(x): 12333, then call PALINDROME(x)
Output : Palindrome : NO

Can you give us some idea on how to solve this program?
Thank you!

[Aakanaar]

if it would take a string then reversing the numbers for comparison would be easy. However, since it takes a long.. i'm not sure how to convert from int back into a string..

the only thing I can think of, is to use a bunch of modulo by 10's, or somethign similar to get the individual digits.

Let me experiment with it and see what I can come up with.

[mikeb570]

You can use stringstream to convert int's to strings or strings to ints.

[Aakanaar]

as soon as I hit submit, it came to mind. I have yet to try it yet, however.

but something along the line of

long original; // original number, i'll skip the step of putting a number into it, for this example.

vector<int> digits; // to hold the digits

while (original >= 10) // while more than 1 digit
{
   int temp = original % 10; // get the last digit
   digits.push_back(temp); // add it to vector
   original = (original - temp) / 10; // get rid of last digit, and shift one
}
digits.push_back(original); // put last digit into vector 

This is untested, but should seperate each digit into a vector, from which you can then do as you see fit, including reversing the order. I would make a copy, reversing the order in the process, then compare the two vectors.

Or, you can, after reversing the order, put it back together as thus:

long reversed; // to put the reversed number in
for (int i = 0; i < digits.size(); i++)
{
   reversed *= 10; // shift everything to the left one
   reversed += digits.at(i); // add new digit to end
}

Give that a try

[helios]

Yes, you can do that...
Or you can do it the fun way!

#include <cstdio>
#include <cmath>

bool palindrome(int n){
	int l=int(log((double)n)/log(10))+1;
	char *temp=new char[l+1];
	fprintf(temp,"%d",n);
	...
}