Write a program to reverse digits of a number
Last Updated :
21 May, 2025
Given an Integer n, find the reverse of its digits.
Examples:
Input: n = 122
Output: 221
Explanation: By reversing the digits of number, number will change into 221.
Input: n = 200
Output: 2
Explanation: By reversing the digits of number, number will change into 2.
Input: n = 12345
Output: 54321
Explanation: By reversing the digits of number, number will change into 54321.
Reversing Digit by Digit
This idea for this approach is to repeatedly extract the last digit of n using the modulus operator (n % 10) and appending it to the reverse number (revNum). After extracting the digit, the number n is reduced by dividing it by 10 (n = n / 10). This process continues until n becomes 0. Finally, the reversed number (revNum) is returned.
C++
#include <bits/stdc++.h>
using namespace std;
int reverseDigits(int n) {
int revNum = 0;
while (n > 0) {
revNum = revNum * 10 + n % 10;
n = n / 10;
}
return revNum;
}
int main() {
int n = 4562;
cout << reverseDigits(n);
return 0;
}
#include <stdio.h>
int reverseDigits(int n) {
int revNum = 0;
while (n > 0) {
revNum = revNum * 10 + n % 10;
n = n / 10;
}
return revNum;
}
int main() {
int n = 4562;
printf("%d",reverseDigits(n));
getchar();
return 0;
}
// Java program to reverse a number
class GfG {
static int reverseDigits(int n) {
int revNum = 0;
while (n > 0) {
revNum = revNum * 10 + n % 10;
n = n / 10;
}
return revNum;
}
public static void main(String[] args) {
int num = 4562;
System.out.println(reverseDigits(num));
}
}
n = 4562
rev = 0
while(n > 0):
a = n % 10
rev = rev * 10 + a
n = n // 10
print(rev)
// C# program to reverse a number
using System;
class GfG {
static int reverseDigits(int n) {
int revNum = 0;
while (n > 0) {
revNum = revNum * 10 + n % 10;
n = n / 10;
}
return revNum;
}
public static void Main() {
int num = 4562;
Console.Write(reverseDigits(num));
}
}
let num = 4562;
function reverseDigits(n) {
let revNum = 0;
while(n > 0)
{
revNum = revNum * 10 + n % 10;
n = Math.floor(n / 10);
}
return revNum;
}
// function call
console.log(reverseDigits(num));
Time Complexity - O(log n)
Space Complexity - O(1)
Using Recursion
This approach uses a recursive function reverseDigits called with the number n and an accumulator variable revNum to store the reversed number, and basePos to keep track of the place value (ones, tens, hundreds, etc.). In each recursive call, the function processes the current digit by dividing n by 10 to reduce it and adding the last digit (n % 10) to revNum, scaled by the basePos. The basePos is then multiplied by 10 to move to the next place value.
C++
#include <bits/stdc++.h>
using namespace std;
void reverseDigits(int n, int &revNum, int &basePos) {
if (n > 0) {
reverseDigits(n / 10, revNum, basePos);
revNum += (n % 10) * basePos;
basePos *= 10;
}
}
int main() {
int n = 4562;
int revNum = 0;
int basePos = 1;
reverseDigits(n, revNum, basePos);
cout << revNum;
return 0;
}
#include <stdio.h>
void reverseDigits(int n, int *revNum, int *basePos) {
if (n > 0) {
reverseDigits(n / 10, revNum, basePos);
*revNum += (n % 10) * (*basePos);
*basePos *= 10;
}
}
int main() {
int n = 4562;
int revNum = 0;
int basePos = 1;
reverseDigits(n, &revNum, &basePos);
printf("%d", revNum);
return 0;
}
class Solution {
// Function to reverse digits recursively
public void reverseDigits(int n, int[] revNum, int[] basePos) {
if (n > 0) {
reverseDigits(n / 10, revNum, basePos);
// Add current digit with its base position
revNum[0] += (n % 10) * basePos[0];
basePos[0] *= 10;
}
}
public static void main(String[] args) {
Solution solution = new Solution();
int n = 4562;
int[] revNum = {0};
int[] basePos = {1};
solution.reverseDigits(n, revNum, basePos);
System.out.println(revNum[0]);
}
}
def reverseDigits(n, revNum, basePos):
if n > 0:
reverseDigits(n // 10, revNum, basePos)
revNum[0] += (n % 10) * basePos[0]
basePos[0] *= 10
# Driver code
n = 4562
revNum = [0]
basePos = [1]
reverseDigits(n, revNum, basePos)
print(revNum[0])
// C# program to reverse digits of a number
using System;
class GfG {
static int reverseDigits(int n) {
int revNum = 0;
int basepos = 1;
if (n > 0) {
reverseDigits(n / 10);
revNum += (n % 10) * basepos;
basepos *= 10;
}
return revNum;
}
public static void Main() {
int n = 4562;
Console.WriteLine(reverseDigits(n));
}
}
class Solution {
// Function to reverse digits recursively
reverseDigits(n, revNum, basePos) {
if (n > 0) {
this.reverseDigits(Math.floor(n / 10), revNum, basePos);
revNum[0] += (n % 10) * basePos[0];
basePos[0] *= 10;
}
}
}
// Driver code
let solution = new Solution();
let n = 4562;
let revNum = [0];
let basePos = [1];
solution.reverseDigits(n, revNum, basePos);
console.log(revNum[0]);
Time Complexity - O(log n)
Space Complexity - O(log n)
This approach reverses a number by converting it into a string, reversing the string, and then converting it back into an integer. This avoids manual digit manipulation by leveraging string operations. The string reversal is done using a built-in function, and the result is then converted back to an integer and returned. This method is straightforward but requires extra space for the string conversion.
C++
// C++ program to reverse a number
#include <bits/stdc++.h>
using namespace std;
int reverseDigits(int n) {
// converting number to string
string s = to_string(n);
// reversing the string
reverse(s.begin(), s.end());
// converting string to integer
n = stoi(s);
// returning integer
return n;
}
int main() {
int n = 4562;
cout << reverseDigits(n);
return 0;
}
// C program to reverse a number
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void reverse(char* begin, char* end) {
char temp;
while (begin < end) {
temp = *begin;
*begin++ = *end;
*end-- = temp;
}
}
void reverseWords(char* s) {
char* word_begin = s;
char* temp = s;
// Reversing individual words as explained in the first step
while (*temp) {
temp++;
if (*temp == '\0') {
reverse(word_begin, temp - 1);
}
else if (*temp == ' ') {
reverse(word_begin, temp - 1);
word_begin = temp + 1;
}
}
reverse(s, temp - 1);
}
int reverseDigits(int n) {
char strin[100];
sprintf(strin, "%d", n);
// reversing the string
reverseWords(strin);
// converting string to integer
n = atoi(strin);
return n;
}
int main() {
int n = 123456;
printf("%d\n",reverseDigits(n));
return 0;
}
// Java program to reverse a number
class GfG {
static int reversDigits(int n) {
// converting number to string
StringBuffer s
= new StringBuffer(String.valueOf(n));
// reversing the string
s.reverse();
// converting string to integer
n = Integer.parseInt(String.valueOf(s));
// returning integer
return n;
}
public static void main(String[] args) {
int n = 4562;
System.out.println(reversDigits(n));
}
}
# Python 3 program to reverse a number
def reversDigits(n):
# converting number to string
s = str(n)
# reversing the string
s = list(s)
s.reverse()
s = ''.join(s)
# converting string to integer
n = int(s)
return n
if __name__ == "__main__":
num = 4562
print(reversDigits(num))
// C# program to reverse a number
using System;
public class GfG {
static string ReverseString(string s) {
char[] array = s.ToCharArray();
Array.Reverse(array);
return new string(array);
}
static int reversDigits(int n) {
// converting number to string
string s = n.ToString();
// reversing the string
s = ReverseString(s);
// converting string to integer
n = int.Parse(s);
// returning integer
return n;
}
static public void Main() {
int n = 4562;
Console.Write(reversDigits(n));
}
}
// Javascript program to reverse a number
function reversDigits(n) {
// converting number to string
let s = n.toString().split("").reverse().join("");
// converting string to integer
n = parseInt(s);
// returning integer
return s;
}
// Driver Code
let n = 4562;
console.log(reversDigits(n));
Time Complexity - O(log n)
Space Complexity - O(1)
Note: The above program doesn't consider leading zeroes. For example, for 100 programs will print 1. If you want to print 001
Using String and Slicing in Python
The approach used is "Using Slicing". This technique involves converting the number into a string, then reversing that string by using slicing operations. After reversing, the string is converted back into a number. This method is simple and efficient but requires additional space for storing the string representation of the number.
Python
# Python 3 program to reverse a number
def reversDigits(n):
# converting number to string
s = str(n)
# reversing the string
s = s[::-1]
# converting string to integer
n = int(s)
# returning integer
return n
if __name__ == "__main__":
n = 4562
print(reversDigits(n))
Time Complexity - O(n)
Space Complexity - O(n)
C++ Program to Reverse a Number
C++ Program to Reverse a Number
Python Program to Reverse any Number
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