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. If you want to print 001 then see this comment from Maheshwar.

Try extensions of above functions that should also work for floating point numbers.

         

Related Tpoics: