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

• Difficulty Level : Easy
• Last Updated : 10 Aug, 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.

Implementation:

## C++

 // C++ program to implement Stack// using linked list so that reverse// can be done with O(1) extra space.#includeusing 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 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 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#

 // 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



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.

My Personal Notes arrow_drop_up