Program To Reverse a Number

ITERATIVE WAY 

Algorithm:

Input:  num
(1) Initialize rev_num = 0
(2) Loop while num > 0
     (a) Multiply rev_num by 10 and add remainder of num  
          divide by 10 to rev_num
               rev_num = rev_num*10 + num%10;
     (b) Divide num by 10
(3) Return rev_num
Example: 

num = 4562 
rev_num = 0
rev_num = rev_num *10 + num%10 = 2 
num = num/10 = 456
rev_num = rev_num *10 + num%10 = 20 + 6 = 26 
num = num/10 = 45
rev_num = rev_num *10 + num%10 = 260 + 5 = 265 
num = num/10 = 4
rev_num = rev_num *10 + num%10 = 2650 + 4 = 2654 
num = num/10 = 0
---------------------------------------------------------------------------

// C program to implement
// the above approach
#include <stdio.h>

// Iterative function to
// reverse digits of num
int reverseDigits(int num)
{
	int rev_num = 0;
	while (num > 0)
	{
		rev_num = rev_num * 10 + num % 10;
		num = num / 10;
	}
	return rev_num;
}

// Driver code
int main()
{
	int num = 4562;
	printf("Reverse of no. is %d",
			reverseDigits(num));

	getchar();
	return 0;
}

------------------------------------
RECURSIVE WAY :
------------------------------------
// C program to reverse digits
// of a number
#include <stdio.h>;

// Recursive function to
// reverse digits of num
int reversDigits(int num)
{
	static int rev_num = 0;
	static int base_pos = 1;
	if (num > 0)
	{
		reversDigits(num / 10);
		rev_num += (num % 10) * base_pos;
		base_pos *= 10;
	}
	return rev_num;
}

// Driver code
int main()
{
	int num = 4562;
	printf("Reverse of no. is %d",
			reversDigits(num));

	getchar();
	return 0;
}