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++
#include <bits/stdc++.h>
using namespace std;
stack <int> st;
void push_digits(int number)
{
while (number != 0)
{
st.push(number % 10);
number = number / 10;
}
}
int reverse_number(int number)
{
push_digits(number);
int reverse = 0;
int i = 1;
while (!st.empty())
{
reverse = reverse + (st.top() * i);
st.pop();
i = i * 10;
}
return reverse;
}
int main()
{
int number = 39997;
cout << reverse_number(number);
return 0;
}
|
Java
import java.util.Stack;
public class GFG
{
static Stack<Integer> st= new Stack<>();
static void push_digits(int number)
{
while(number != 0)
{
st.push(number % 10);
number = number / 10;
}
}
static int reverse_number(int number)
{
push_digits(number);
int reverse = 0;
int i = 1;
while (!st.isEmpty())
{
reverse = reverse + (st.peek() * i);
st.pop();
i = i * 10;
}
return reverse;
}
public static void main(String[] args)
{
int number = 39997;
System.out.println(reverse_number(number));
}
}
|
Python3
st = [];
def push_digits(number):
while (number != 0):
st.append(number % 10);
number = int(number / 10);
def reverse_number(number):
push_digits(number);
reverse = 0;
i = 1;
while (len(st) > 0):
reverse = reverse + (st[len(st) - 1] * i);
st.pop();
i = i * 10;
return reverse;
number = 39997;
print(reverse_number(number));
|
C#
using System;
using System.Collections.Generic;
class GFG
{
public static Stack<int> st = new Stack<int>();
public static void push_digits(int number)
{
while (number != 0)
{
st.Push(number % 10);
number = number / 10;
}
}
public static int reverse_number(int number)
{
push_digits(number);
int reverse = 0;
int i = 1;
while (st.Count > 0)
{
reverse = reverse + (st.Peek() * i);
st.Pop();
i = i * 10;
}
return reverse;
}
public static void Main(string[] args)
{
int number = 39997;
Console.WriteLine(reverse_number(number));
}
}
|
PHP
<?php
$st = array();
function push_digits($number)
{
global $st;
while ($number != 0)
{
array_push($st, $number % 10);
$number = (int)($number / 10);
}
}
function reverse_number($number)
{
global $st;
push_digits($number);
$reverse = 0;
$i = 1;
while (!empty($st))
{
$reverse = $reverse +
($st[count($st) - 1] * $i);
array_pop($st);
$i = $i * 10;
}
return $reverse;
}
$number = 39997;
echo reverse_number($number);
?>
|
Javascript
<script>
let st = [];
function push_digits(number)
{
while (number != 0)
{
st.push(number % 10);
number = Math.floor(number / 10);
}
}
function reverse_number(number)
{
push_digits(number);
let reverse = 0;
let i = 1;
while (st.length != 0) {
reverse = reverse + (st[st.length - 1] * i);
st.pop();
i = i * 10;
}
return reverse;
}
let number = 39997;
document.write(reverse_number(number));
</script>
|
Time Complexity: O( logN )
Auxiliary Space: O( logN ), Where N is the input number.
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