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Reverse a stack without using extra space in O(n)

Last Updated : 10 Jan, 2023
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Reverse a Stack without using recursion and extra space. Even the functional Stack is not allowed.

Examples:  

Input : 1->2->3->4
Output : 4->3->2->1

Input :  6->5->4
Output : 4->5->6

We have discussed a way of reversing a stack in the below post.
Reverse a Stack using Recursion

The above solution requires O(n) extra space. We can reverse a stack in O(1) time if we internally represent the stack as a linked list. Reverse a stack would require a reversing of a linked list which can be done with O(n) time and O(1) extra space.
Note that push() and pop() operations still take O(1) time.  

Implementation:

C++




// C++ program to implement Stack 
// using linked list so that reverse
// can be done with O(1) extra space.
#include<bits/stdc++.h>
using namespace std;
  
class StackNode {
    public:
    int data;
    StackNode *next;
      
    StackNode(int data)
    {
        this->data = data;
        this->next = NULL;
    }
};
  
class Stack {
      
    StackNode *top;
      
    public:
      
    // Push and pop operations
    void push(int data)
    {
        if (top == NULL) {
            top = new StackNode(data);
            return;
        }
        StackNode *s = new StackNode(data);
        s->next = top;
        top = s;
    }
      
    StackNode* pop()
    {
        StackNode *s = top;
        top = top->next;
        return s;
    }
  
    // prints contents of stack
    void display()
    {
        StackNode *s = top;
        while (s != NULL) {
            cout << s->data << " ";
            s = s->next;
        }
        cout << endl;
    }
  
    // Reverses the stack using simple
    // linked list reversal logic.
    void reverse()
    {
        StackNode *prev, *cur, *succ;
        cur = prev = top;
        cur = cur->next;
        prev->next = NULL;
        while (cur != NULL) {
  
            succ = cur->next;
            cur->next = prev;
            prev = cur;
            cur = succ;
        }
        top = prev;
    }
};
  
// driver code
int main()
{
    Stack *s = new Stack();
    s->push(1);
    s->push(2);
    s->push(3);
    s->push(4);
    cout << "Original Stack" << endl;;
    s->display();
    cout << endl;
      
    // reverse
    s->reverse();
  
    cout << "Reversed Stack" << endl;
    s->display();
      
    return 0;
}
// This code is contributed by Chhavi.


Java




// Java program to implement Stack using linked
// list so that reverse can be done with O(1) 
// extra space.
class StackNode {
    int data;
    StackNode next;
    public StackNode(int data)
    {
        this.data = data;
        this.next = null;
    }
}
  
class Stack {
    StackNode top;
  
    // Push and pop operations
    public void push(int data)
    {
        if (this.top == null) {
            top = new StackNode(data);
            return;
        }
        StackNode s = new StackNode(data);
        s.next = this.top;
        this.top = s;
    }
    public StackNode pop()
    {
        StackNode s = this.top;
        this.top = this.top.next;
        return s;
    }
  
    // prints contents of stack
    public void display()
    {
        StackNode s = this.top;
        while (s != null) {
            System.out.print(s.data + " ");
            s = s.next;
        }
        System.out.println();
    }
  
    // Reverses the stack using simple
    // linked list reversal logic.
    public void reverse()
    {
        StackNode prev, cur, succ;
        cur = prev = this.top;
        cur = cur.next;
        prev.next = null;
        while (cur != null) {
  
            succ = cur.next;
            cur.next = prev;
            prev = cur;
            cur = succ;
        }
        this.top = prev;
    }
}
  
public class reverseStackWithoutSpace {
    public static void main(String[] args)
    {
        Stack s = new Stack();
        s.push(1);
        s.push(2);
        s.push(3);
        s.push(4);
        System.out.println("Original Stack");
        s.display();
  
        // reverse
        s.reverse();
  
        System.out.println("Reversed Stack");
        s.display();
    }
}


Python3




# Python3 program to implement Stack 
# using linked list so that reverse
# can be done with O(1) extra space.
class StackNode:
      
    def __init__(self, data):
          
        self.data = data
        self.next = None
  
class Stack:
      
    def __init__(self):
           
        self.top = None
       
    # Push and pop operations
    def push(self, data):
      
        if (self.top == None):
            self.top = StackNode(data)
            return
          
        s = StackNode(data)
        s.next = self.top
        self.top = s
       
    def pop(self):
      
        s = self.top
        self.top = self.top.next
        return s
   
    # Prints contents of stack
    def display(self):
      
        s = self.top
          
        while (s != None):
            print(s.data, end = ' ')
            s = s.next
          
    # Reverses the stack using simple
    # linked list reversal logic.
    def reverse(self):
  
        prev = self.top
        cur = self.top
        cur = cur.next
        succ = None
        prev.next = None
          
        while (cur != None):
            succ = cur.next
            cur.next = prev
            prev = cur
            cur = succ
          
        self.top = prev
      
# Driver code
if __name__=='__main__':
      
    s = Stack()
    s.push(1)
    s.push(2)
    s.push(3)
    s.push(4)
      
    print("Original Stack")
    s.display()
    print()
       
    # Reverse
    s.reverse()
   
    print("Reversed Stack")
    s.display()
       
# This code is contributed by rutvik_56


C#




// C# program to implement Stack using linked
// list so that reverse can be done with O(1) 
// extra space.
using System; 
  
public class StackNode
{
    public int data;
    public StackNode next;
    public StackNode(int data)
    {
        this.data = data;
        this.next = null;
    }
}
  
public class Stack 
{
    public StackNode top;
  
    // Push and pop operations
    public void push(int data)
    {
        if (this.top == null)
        {
            top = new StackNode(data);
            return;
        }
        StackNode s = new StackNode(data);
        s.next = this.top;
        this.top = s;
    }
      
    public StackNode pop()
    {
        StackNode s = this.top;
        this.top = this.top.next;
        return s;
    }
  
    // prints contents of stack
    public void display()
    {
        StackNode s = this.top;
        while (s != null
        {
            Console.Write(s.data + " ");
            s = s.next;
        }
        Console.WriteLine();
    }
  
    // Reverses the stack using simple
    // linked list reversal logic.
    public void reverse()
    {
        StackNode prev, cur, succ;
        cur = prev = this.top;
        cur = cur.next;
        prev.next = null;
        while (cur != null)
        {
            succ = cur.next;
            cur.next = prev;
            prev = cur;
            cur = succ;
        }
        this.top = prev;
    }
}
  
public class reverseStackWithoutSpace 
{
    // Driver code
    public static void Main(String []args)
    {
        Stack s = new Stack();
        s.push(1);
        s.push(2);
        s.push(3);
        s.push(4);
        Console.WriteLine("Original Stack");
        s.display();
  
        // reverse
        s.reverse();
  
        Console.WriteLine("Reversed Stack");
        s.display();
    }
}
  
// This code is contributed by Arnab Kundu


Javascript




<script>
  
// JavaScript program to implement Stack 
// using linked list so that reverse can 
// be done with O(1) extra space.
class StackNode 
{
    constructor(data) 
    {
        this.data = data;
        this.next = null;
    }
}
  
class Stack 
{
    top = null;
      
    // Push and pop operations
    push(data)
    {
        if (this.top == null
        {
            this.top = new StackNode(data);
            return;
        }
        var s = new StackNode(data);
        s.next = this.top;
        this.top = s;
    }
      
    pop() 
    {
        var s = this.top;
        this.top = this.top.next;
        return s;
    }
      
    // Prints contents of stack
    display() 
    {
        var s = this.top;
        while (s != null)
        {
            document.write(s.data + " ");
            s = s.next;
        }
        document.write("<br>");
    }
      
    // Reverses the stack using simple
    // linked list reversal logic.
    reverse() 
    {
        var prev, cur, succ;
        cur = prev = this.top;
        cur = cur.next;
        prev.next = null;
          
        while (cur != null
        {
            succ = cur.next;
            cur.next = prev;
            prev = cur;
            cur = succ;
        }
        this.top = prev;
    }
}
  
// Driver code
var s = new Stack();
s.push(1);
s.push(2);
s.push(3);
s.push(4);
document.write("Original Stack <br>");
s.display();
  
// Reverse
s.reverse();
  
document.write("Reversed Stack <br>");
s.display();
  
// This code is contributed by rdtank
  
</script>


Output

Original Stack
4 3 2 1 

Reversed Stack
1 2 3 4 

Time Complexity: O(n), as we are using a loop to traverse n times. Where n is the number of nodes in the linked list.
Auxiliary Space: O(1), as we are not using any extra space.

 



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