Write a program to reverse digits of a number

Write a program to reverse digits of an integer.

Examples :

Input : num = 12345
Output : 54321

Input : num = 876
Output : 678

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

Flowchart:

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:

C++

#include <bits/stdc++.h> 
  
using namespace std; 
/* 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; 
    cout << "Reverse of no. is "
         << reversDigits(num); 
    getchar(); 
    return 0; 
} 
  
// This code is contributed 
// by Akanksha Rai(Abby_akku) 

C

#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; 
} 

Java

// Java program to reverse a number  
  
class GFG 
{ 
    /* Iterative function to reverse 
    digits of num*/
    static 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 code 
    public static void main (String[] args)  
    { 
        int num = 4562; 
        System.out.println("Reverse of no. is " 
                           + reversDigits(num)); 
    } 
} 
  
// This code is contributed by Anant Agarwal. 

Python

# Python program to reverse a number  
  
n = 4562; 
rev = 0
  
while(n > 0): 
    a = n % 10
    rev = rev * 10 + a 
    n = n // 10
      
print(rev) 
  
# This code is contributed by Shariq Raza  

C#

// C# program to reverse a number  
using System; 
  
class GFG 
{ 
    // Iterative function to  
    // reverse digits of num 
    static 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 code 
    public static void Main()  
    { 
        int num = 4562; 
        Console.Write("Reverse of no. is "
                        + reversDigits(num)); 
    } 
} 
  
// This code is contributed by Sam007 

PHP

<?php 
// Iterative function to  
// reverse digits of num 
function reversDigits($num) 
{ 
    $rev_num = 0; 
    while($num > 1) 
    { 
        $rev_num = $rev_num * 10 +  
                        $num % 10; 
        $num = (int)$num / 10; 
    } 
    return $rev_num; 
} 
  
// Driver Code 
$num = 4562; 
echo "Reverse of no. is ",  
       reversDigits($num); 
  
// This code is contributed by aj_36 
?> 

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

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

C++

// C++ program to reverse digits of a number 
#include <bits/stdc++.h> 
using namespace std; 
/* 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 Code 
int main() 
{ 
    int num = 4562; 
    cout << "Reverse of no. is " 
         << reversDigits(num); 
  
    return 0; 
} 
  
// This code is contributed  
// by Akanksha Rai(Abby_akku) 

C

// C program to reverse digits of a number 
#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; 
} 

Java

// Java program to reverse digits of a number 
  
// Recursive function to  
// reverse digits of num 
class GFG 
{ 
static int rev_num = 0; 
static int base_pos = 1; 
static int reversDigits(int num) 
{ 
    if(num > 0) 
    { 
        reversDigits(num / 10); 
        rev_num += (num % 10) * base_pos; 
        base_pos *= 10; 
    } 
return rev_num; 
} 
  
// Driver Code 
public static void main(String[] args) 
{ 
    int num = 4562; 
    System.out.println(reversDigits(num)); 
} 
} 
  
// This code is contributed by mits 

Python3

# Python 3 program to reverse digits 
# of a number 
rev_num = 0
base_pos = 1
  
# Recursive function to reverse 
# digits of num 
def reversDigits(num): 
    global rev_num 
    global base_pos 
    if(num > 0): 
        reversDigits((int)(num / 10)) 
        rev_num += (num % 10) * base_pos 
        base_pos *= 10
    return rev_num 
      
# Driver Code  
num = 4562
print("Reverse of no. is ", 
         reversDigits(num)) 
      
# This code is contributed by Rajput-Ji 

C#

// C# program to reverse digits of a number 
  
// Recursive function to  
// reverse digits of num 
using System; 
class GFG 
{ 
static int rev_num = 0; 
static int base_pos = 1; 
static int reversDigits(int num) 
{ 
    if(num > 0) 
    { 
        reversDigits(num / 10); 
        rev_num += (num % 10) * base_pos; 
        base_pos *= 10; 
    } 
return rev_num; 
} 
  
// Driver Code 
public static void Main() 
{ 
    int num = 4562; 
    Console.WriteLine(reversDigits(num)); 
} 
} 
  
// This code is contributed  
// by inder_verma 

PHP

<?php 
// PHP program to reverse digits of a number 
$rev_num = 0;  
$base_pos = 1; 
  
/* Recursive function to  
reverse digits of num*/
function reversDigits($num)  
{  
    global $rev_num;  
    global $base_pos;  
    if($num > 0)  
    {  
        reversDigits((int)($num / 10));  
        $rev_num += ($num % 10) *  
                     $base_pos;  
        $base_pos *= 10;  
    }  
    return $rev_num;  
}  
  
// Driver Code 
$num = 4562;  
echo "Reverse of no. is ",  
       reversDigits($num);  
  
// This code is contributed by ajit 
?> 

Output:

Reverse of no. is 2654

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

Reverse digits of an integer with overflow handled

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.

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My Personal Notes

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

C++

C

Java

Python

C#

PHP

C++

C

Java

Python3

C#

PHP

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