Program to Reverse a Number using Recursion

The program uses recursion to reverse an input number. A function is recursively called to extract each digit from the number and place it in reverse order. The final output is a reversed number.

Problem Definition

The program receives an input number and reverse each by extracting the digits one at a time from original number n. It starts with two initial conditions. The remainder r = 0 and if n=0 then return 0.

The procedure to reverse the number 567 is shown below. The program does the following tasks to reverse the input number.

1. Extract a digit from the unit position.
2. Each iteration multiply the number with 10 to reverse it.
3. Next extracted digit is added to previously obtained result.
4. After final extraction the number is completely reversed.

Go through each step below to understand the working of the reverse() function.

Step 1: Extract the number 7

\begin{aligned}&r = r + n \hspace{2mm}\% \hspace{2mm}10\\ \\
&r = 0 + 567 \hspace{2mm}\% \hspace{2mm}10\\ \\ 
&t = 7
\end{aligned}

Step 2: Change the position of a number from unit place to tenth place.

\begin{aligned}
&r = r * 10\\ \\
& r = 7 * 10\\ \\
 &r = 70
\end{aligned}

Step 3: Get ready for the next extraction.

\begin{aligned}
&= reverse (n/10)\\ \\
&= reverse (567/10)\\ \\
&= reverse (56)
\end{aligned}

Step 4: Extraction the second digit when value of r = 70 from Step 2.

\begin{aligned}
&r = 70 + n \% 10\\ \\
&r = 70 + 56 \% 10\\ \\
&r = 70 + 6\\ \\
&r = 76
\end{aligned}

Step 5: Change position of r to tenth place to hundredth place.

\begin{aligned} &r = r * 10\\ \\
&r = 76 * 10\\ \\
&r = 760
\end{aligned}

Step 6: Get ready for a new n value.

\begin{aligned}
&= reverse (n/10)\\ \\
& = reverse ( 56/1)\\ \\
& = reverse (5)
\end{aligned}

Step 7: Extract last digit from original number, when r = 760

\begin{aligned}
&r = r + n \% 10\\ \\ 
& r = 760 + 5\\ \\
&r = 765
\end{aligned}

The number is successfully reversed.

Note:- The above steps are demonstration of reversing a 3 digit number, hence, the number of steps are limited. If you take a larger number there will be more steps.

Flowchart – reverse a number using recursion

Program Codes – reverse a number using recursion

#include 
#include 
int reverse(int);
int main()
{
        int n, r, i;
        printf("Enter a Number to Reverse:");
        scanf("%d", & n);
        r = reverse(n);
        for (i = 0; i < 30; i++)
            printf("_"); printf("\n\n");
            printf("%d\n\n", r);
        for (i = 0; i < 30; i++)
            printf("_"); printf("\n\n");
        system("PAUSE");
        return (0);
}
/* The Reverse Function */
int reverse(int n)
{
        static long r = 0;
        if (n == 0)
        {
            return 0;
        }
        r = r * 10;
        r = r + n % 10;
        reverse(n / 10);
        return r;
}

#include <iostream>
using namespace std;

long reverse(int n)
{
    static long r = 0;   // static variable to retain value
    if (n == 0)
        return 0;

    r = r * 10 + (n % 10);
    reverse(n / 10);
    return r;
}

int main()
{
    int n;
    long r;

    cout << "Enter a Number to Reverse: ";
    cin >> n;

    r = reverse(n);

    for (int i = 0; i < 30; i++)
        cout << "_";
    cout << "\n\n";

    cout << r << "\n\n";

    for (int i = 0; i < 30; i++)
        cout << "_";
    cout << "\n";

    return 0;
}

import java.util.Scanner;

class ReverseNumber {

    static long r = 0;   // static variable

    static long reverse(int n) {
        if (n == 0) {
            return 0;
        }
        r = r * 10 + (n % 10);
        reverse(n / 10);
        return r;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        System.out.print("Enter a Number to Reverse: ");
        int n = sc.nextInt();

        long result = reverse(n);

        for (int i = 0; i < 30; i++)
            System.out.print("_");
        System.out.println("\n");

        System.out.println(result + "\n");

        for (int i = 0; i < 30; i++)
            System.out.print("_");
        System.out.println();

        sc.close();
    }
}

def reverse(n, r=0):
    if n == 0:
        return r
    return reverse(n // 10, r * 10 + n % 10)


n = int(input("Enter a Number to Reverse: "))
result = reverse(n)

print("_" * 30)
print("\n", result, "\n")
print("_" * 30)

Output

The program ask for an input number, when you enter a number – for example, 244. The program reverses the digits of the number and prints on the console. The solution is 442.

Enter a Number to Reverse: 244
_______________________________
442
_______________________________