Finding reverse of a number

METHOD 1:

#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;
}

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

    getchar();
    return 0;
}

OUTPUT:
Reverse of no. is 2654

Time complexity: O(log(n)) where n is the input number.

The above program gives the output 1 for an input 100. To get the output as 001, a small addition is done in the above program.


int reversDigits(int num)
{
    int rev_num = 0,zero=0;
    while(num>0)
    {
        if(num%10==0)    
        {
            zero++;
            num=num/10;
        }
        else    break;
    }
    while(num > 0)
    {
        rev_num = rev_num*10 + num%10;
        num = num/10;
    }
    return rev_num;
}



METHOD 2:

#include <stdio.h>
#include <conio.h>

int main()
{
    char a[5];     //Get the number as a string
    int n;
    printf("enter a number\n");
    scanf("%s",a);
    strrev(a);      //reverse the string using strrev() function
    n=atoi(a);     //convert string to integer
    printf("%d",n);
    getch();
    return 0;
}

OUTPUT:
enter a number
1234
4321