# Reverse a number using stack in C++

We are given an integer number Num as input. The goal is to find the reverse of the number using stack.

Stack:- A stack is a data structure in C++ which stores data in LIFO ( Last in First Out) manner. Major operations of stack are-:

Declaration-: stack <int> stck; //stck is now a stack variable.

• Finding Top using top(). Function stck.top() returns reference of top element in the stck

• Removing Top using pop(). Function removes topmost element from the stck

• Adding element to top using push(). Function stck.push( value ) adds item value in stack. Value should be of type stck.

• Check if staxk is empty using empty(). Function stck.empty() returns true if stack is empty.

## Examples

Input − Num = 33267

Output − Reverse of number is: 76233

Explanation

First we will push all elements to stack

7 - 6 - 2 - 3 - 3 ← top

7 * 10000 + 6 * 1000 + 2*100 + 3*10 + 3*1 ←

= 70000 + 6000 + 200 + 30 + 3 ←

= 76233

Input − Num = 111000

Output − Reverse of number is: 111

Explanation

First we will push all elements to stack

0 - 0 - 0 - 1 - 1 - 1 ← top

0 * 100000 + 0 * 10000 + 0*1000 + 1*100 + 1*10 + 1*1 ←

= 0 + 0 + 0 + 100 + 10 + 1 ←

= 111

## Approach used in the below program is as follows

In this approach we will first take remainders of input number and push to stack and reduce number by 10 until number becomes 0. In this way stack will be filled with top as first digit.

• Take the input number Num.

• Take empty stack for integers using stack<int> stck.

• Function pushDigts(int num1) takes num1 and adds it to stack with first digit on top.

• Take rem as variable.

• Using a while loop check if num1 is non-zero, if true then set rem=num1%10.

• Push rem to stack.

• Reduce num1 by 10 for 2nd digit and so on.

• Now reverse the number using elements of stack with function revrseNum().

• Take variables revrs, topp, temp, i.

• While the stack is not empty

• Take the topmost element as topp=stck.top().

• Reduce stack using stck.pop().

• Set temp=topp*i.

• Increase i by i*10 in multiples of 100.

• At the end return the reverse of the input num as revrs.

• Print result obtained inside main.

## Example

#include <bits/stdc++.h>
using namespace std;
stack <int> stck;
void pushDigts(int num1){
int rem;
while (num1 > 0){
rem=num1 % 10;
stck.push(rem);
num1 = num1 / 10;
}
}
int revrseNum(){
int revrs = 0;
int i = 1;
int temp;
int topp;
while (!stck.empty()){
topp=stck.top();
stck.pop();
temp=topp*i;
revrs = revrs + temp;
i *= 10;
}
return revrs;
}
int main(){
int Num = 43556;
pushDigts(Num);
cout<<"Reverse of number is: "<<revrseNum();
return 0;
}

## Output

If we run the above code it will generate the following Output

Reverse of number is: 65534