# Reverse a number using stack

• Difficulty Level : Easy
• Last Updated : 23 Jul, 2021

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

```Input : 365
Output : 563

Input : 6899
Output : 9986```

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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 poping 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

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## Javascript

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Output:

`79993`

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

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