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

24 Jun, 2022
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.  
C++
Java
Python3
C#
Javascript
// 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 codeint 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 contribute by Chhavi.
// 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 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 codeif __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# 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
<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 codevar s = new Stack();s.push(1);s.push(2);s.push(3);s.push(4);document.write("Original Stack <br>");s.display();  // Reverses.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.This article is contributed by Niharika Sahai. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.Please write comments if you find anything incorrect, or if you want to share more information about the topic discussed above. 
lakshaygupta2807
andrew1234
rutvik_56
rdtank
riddhijain1826
rohan07
Linked List
Stack
Linked List
Stack
Writing code in comment?
Please use ide.geeksforgeeks.org,
generate link and share the link here.
Linked List | Set 1 (Introduction)
Linked List | Set 2 (Inserting a node)
Reverse a linked list
Linked List | Set 3 (Deleting a node)
LinkedList in Java
Stack in Python
Stack Class in Java
Check for Balanced Brackets in an expression (well-formedness) using Stack
Introduction to Data Structures
Stack | Set 2 (Infix to Postfix)