Reverse digits of number

Write a C program to reverse digits of 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 = 265 + 4 = 2654
num = num/10 = 0

Program:

#include <stdio.h>   /* Iterative function to reverse digits of num*/ int reversDigits(int num) {     int rev_num = 0;     while(num > 0)     {         rev_num = rev_num*10 + num%10;         num = num/10;     }     return rev_num; }   /*Driver program to test reversDigits*/ int main() {     int num = 4562;     printf("Reverse of no. is %d", reversDigits(num));       getchar();     return 0; }

Time Complexity: O(Log(n)) where n is the input number.

RECURSIVE WAY
Thanks to Raj for adding this to the original post.

#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 program to test reversDigits*/ int main() {     int num = 4562;     printf("Reverse of no. is %d", reversDigits(num));       getchar();     return 0; }

Time Complexity: O(Log(n)) where n is the input number

Note that above above program doesn’t consider leading zeroes. For example, for 100 program will print 1.