Reverse a number using stack
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 <bits/stdc++.h>using namespace std;// Stack to maintain order of digitsstack <int> st;// Function to push digits into stackvoid push_digits(int number){ while (number != 0) { st.push(number % 10); number = number / 10; }}// Function to reverse the numberint 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 functionint main(){ int number = 39997; // Function call to reverse number cout << reverse_number(number); return 0;} |
Java
// Java program to reverse the number// using a stackimport java.util.Stack;public class GFG{ // Stack to maintain order of digits static Stack<Integer> 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 digitsst = [];# Function to push digits into stackdef push_digits(number): while (number != 0): st.append(number % 10); number = int(number / 10);# Function to reverse the numberdef 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 Codenumber = 39997;# Function call to reverse numberprint(reverse_number(number));# This code is contributed by mits |
C#
// C# program to reverse the number// using a stackusing System;using System.Collections.Generic;class GFG{// Stack to maintain order of digitspublic static Stack<int> st = new Stack<int>();// Function to push digits into stackpublic static void push_digits(int number){ while (number != 0) { st.Push(number % 10); number = number / 10; }}// Function to reverse the numberpublic 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 Codepublic static void Main(string[] args){ int number = 39997; Console.WriteLine(reverse_number(number));}}// This code is contributed by Shrikant13 |
PHP
<?php// PHP program to reverse the number// using a stack// Stack to maintain order of digits$st = array();// Function to push digits into stackfunction push_digits($number){ global $st; while ($number != 0) { array_push($st, $number % 10); $number = (int)($number / 10); }}// Function to reverse the numberfunction reverse_number($number){ global $st; // Function call to push number's // digits to stack push_digits($number); $reverse = 0; $i = 1; // Popping the digits and forming // the reversed number while (!empty($st)) { $reverse = $reverse + ($st[count($st) - 1] * $i); array_pop($st); $i = $i * 10; } // Return the reversed number formed return $reverse;}// Driver Code$number = 39997;// Function call to reverse numberecho reverse_number($number);// This code is contributed by mits?> |
Javascript
<script> // JavaScript program for the above approach // Stack to maintain order of digits let st = []; // Function to push digits into stack function push_digits(number) { while (number != 0) { st.push(number % 10); number = Math.floor(number / 10); } } // Function to reverse the number function reverse_number(number) { // Function call to push number's // digits to stack push_digits(number); let reverse = 0; let i = 1; // Popping the digits and forming // the reversed number while (st.length != 0) { reverse = reverse + (st[st.length - 1] * i); st.pop(); i = i * 10; } // Return the reversed number formed return reverse; } // Driver program to test above function let number = 39997; // Function call to reverse number document.write(reverse_number(number));// This code is contributed by Potta Lokesh</script> |
Output:
79993
Time Complexity: O( logN )
Auxiliary Space: O( logN ), Where N is the input number.
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