 Open in App
Not now

# Reverse a number using stack

• Difficulty Level : Easy
• Last Updated : 15 Jul, 2022

Given a number , write a program to reverse this number using stack.

Examples:

```Input : 365
Output : 563

Input : 6899
Output : 9986```

We have already discussed the simple method to reverse a number in this post. In this post we will discuss about how to reverse a number using stack.
The idea to do this is to extract digits of the number and push the digits on to a stack. Once all of the digits of the number are pushed to the stack, we will start popping the contents of stack one by one and form a number.
As stack is a LIFO data structure, digits of the newly formed number will be in reverse order.

Below is the implementation of above idea:

## C++

 `// CPP program to reverse the number``// using a stack` `#include ``using` `namespace` `std;` `// Stack to maintain order of digits``stack <``int``> st;` `// Function to push digits into stack``void` `push_digits(``int` `number)``{``    ``while` `(number != 0)``    ``{``        ``st.push(number % 10);``        ``number = number / 10;``    ``}``}` `// Function to reverse the number``int` `reverse_number(``int` `number)``{``    ``// Function call to push number's``    ``// digits to stack``    ``push_digits(number);``    ` `    ``int` `reverse = 0;``    ``int` `i = 1;``    ` `    ``// Popping the digits and forming``    ``// the reversed number``    ``while` `(!st.empty())``    ``{``        ``reverse = reverse + (st.top() * i);``        ``st.pop();``        ``i = i * 10;``    ``}``    ` `    ``// Return the reversed number formed``    ``return` `reverse;``}` `// Driver program to test above function``int` `main()``{``    ``int` `number = 39997;``    ` `    ``// Function call to reverse number``    ``cout << reverse_number(number);``    ` `    ``return` `0;``}`

## Java

 `// Java program to reverse the number``// using a stack``import` `java.util.Stack;` `public` `class` `GFG``{``    ``// Stack to maintain order of digits``    ``static` `Stack st= ``new` `Stack<>();` `    ``// Function to push digits into stack``    ``static` `void` `push_digits(``int` `number)``    ``{``        ``while``(number != ``0``)``        ``{``            ``st.push(number % ``10``);``            ``number = number / ``10``;``        ``}``    ``}` `    ``// Function to reverse the number``    ``static` `int` `reverse_number(``int` `number)``    ``{``        ``// Function call to push number's``        ``// digits to stack``        ``push_digits(number);``        ``int` `reverse = ``0``;``        ``int` `i = ``1``;` `        ``// Popping the digits and forming``        ``// the reversed number``        ``while` `(!st.isEmpty())``        ``{``            ``reverse = reverse + (st.peek() * i);``            ``st.pop();``            ``i = i * ``10``;``        ``}` `        ``// Return the reversed number formed``        ``return` `reverse;``    ``}` `    ``// Driver program to test above function``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `number = ``39997``;``        ``System.out.println(reverse_number(number));``    ``}``}``// This code is contributed by Sumit Ghosh`

## Python3

 `# Python3 program to reverse the``# number using a stack` `# Stack to maintain order of digits``st ``=` `[];` `# Function to push digits into stack``def` `push_digits(number):` `    ``while` `(number !``=` `0``):``        ``st.append(number ``%` `10``);``        ``number ``=` `int``(number ``/` `10``);` `# Function to reverse the number``def` `reverse_number(number):``    ` `    ``# Function call to push number's``    ``# digits to stack``    ``push_digits(number);``    ` `    ``reverse ``=` `0``;``    ``i ``=` `1``;``    ` `    ``# Popping the digits and forming``    ``# the reversed number``    ``while` `(``len``(st) > ``0``):``        ``reverse ``=` `reverse ``+` `(st[``len``(st) ``-` `1``] ``*` `i);``        ``st.pop();``        ``i ``=` `i ``*` `10``;``    ` `    ``# Return the reversed number formed``    ``return` `reverse;` `# Driver Code``number ``=` `39997``;` `# Function call to reverse number``print``(reverse_number(number));` `# This code is contributed by mits`

## C#

 `// C# program to reverse the number``// using a stack``using` `System;``using` `System.Collections.Generic;` `class` `GFG``{``// Stack to maintain order of digits``public` `static` `Stack<``int``> st = ``new` `Stack<``int``>();` `// Function to push digits into stack``public` `static` `void` `push_digits(``int` `number)``{``    ``while` `(number != 0)``    ``{``        ``st.Push(number % 10);``        ``number = number / 10;``    ``}``}` `// Function to reverse the number``public` `static` `int` `reverse_number(``int` `number)``{``    ``// Function call to push number's``    ``// digits to stack``    ``push_digits(number);``    ``int` `reverse = 0;``    ``int` `i = 1;` `    ``// Popping the digits and forming``    ``// the reversed number``    ``while` `(st.Count > 0)``    ``{``        ``reverse = reverse + (st.Peek() * i);``        ``st.Pop();``        ``i = i * 10;``    ``}` `    ``// Return the reversed number formed``    ``return` `reverse;``}` `// Driver Code``public` `static` `void` `Main(``string``[] args)``{``    ``int` `number = 39997;``    ``Console.WriteLine(reverse_number(number));``}``}` `// This code is contributed by Shrikant13`

## PHP

 ``

## Javascript

 ``

Output

`79993`

Time Complexity: O( logN )
Auxiliary Space: O( logN ), Where N is the input number.

This article is contributed by Rohit Thapliyal. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

My Personal Notes arrow_drop_up