Reverse a Linked List

Given a pointer to the head node of a linked list, the task is to reverse the linked list. We need to reverse the list by changing the links between nodes.

Examples: 

Input: Head of following linked list 
1->2->3->4->NULL 
Output: Linked list should be changed to, 
4->3->2->1->NULL

Input: Head of following linked list 
1->2->3->4->5->NULL 
Output: Linked list should be changed to, 
5->4->3->2->1->NULL

Input: NULL 
Output: NULL

Input: 1->NULL 
Output: 1->NULL 

Reverse a linked list by Iterative Method:

The idea is to use three pointers curr, prev, and next to keep track of nodes to update reverse links.

Follow the steps below to solve the problem:

• Initialize three pointers prev as NULL, curr as head, and next as NULL.
• Iterate through the linked list. In a loop, do the following:
  • Before changing the next of curr, store the next node 
    • next = curr -> next
  • Now update the next pointer of curr to the prev
    • curr -> next = prev 
  • Update prev as curr and curr as next
    • prev = curr 
    • curr = next

Below is the implementation of the above approach: 

C++

// Iterative C++ program to reverse a linked list
#include <bits/stdc++.h>
using namespace std;

/* Link list node */
struct Node {
    int data;
    struct Node* next;
    Node(int data)
    {
        this->data = data;
        next = NULL;
    }
};

struct LinkedList {
    Node* head;
    LinkedList() { head = NULL; }

    /* Function to reverse the linked list */
    void reverse()
    {
        // Initialize current, previous and next pointers
        Node* current = head;
        Node *prev = NULL, *next = NULL;

        while (current != NULL) {
            // Store next
            next = current->next;
            // Reverse current node's pointer
            current->next = prev;
            // Move pointers one position ahead.
            prev = current;
            current = next;
        }
        head = prev;
    }

    /* Function to print linked list */
    void print()
    {
        struct Node* temp = head;
        while (temp != NULL) {
            cout << temp->data << " ";
            temp = temp->next;
        }
    }

    void push(int data)
    {
        Node* temp = new Node(data);
        temp->next = head;
        head = temp;
    }
};

/* Driver code*/
int main()
{
    /* Start with the empty list */
    LinkedList ll;
    ll.push(20);
    ll.push(4);
    ll.push(15);
    ll.push(85);

    cout << "Given linked list\n";
    ll.print();

    ll.reverse();

    cout << "\nReversed linked list \n";
    ll.print();
    return 0;
}

C

// Iterative C program to reverse a linked list
#include <stdio.h>
#include <stdlib.h>

/* Link list node */
struct Node {
    int data;
    struct Node* next;
};

/* Function to reverse the linked list */
static void reverse(struct Node** head_ref)
{
    struct Node* prev = NULL;
    struct Node* current = *head_ref;
    struct Node* next = NULL;
    while (current != NULL) {
        // Store next
        next = current->next;

        // Reverse current node's pointer
        current->next = prev;

        // Move pointers one position ahead.
        prev = current;
        current = next;
    }
    *head_ref = prev;
}

/* Function to push a node */
void push(struct Node** head_ref, int new_data)
{
    struct Node* new_node
        = (struct Node*)malloc(sizeof(struct Node));
    new_node->data = new_data;
    new_node->next = (*head_ref);
    (*head_ref) = new_node;
}

/* Function to print linked list */
void printList(struct Node* head)
{
    struct Node* temp = head;
    while (temp != NULL) {
        printf("%d ", temp->data);
        temp = temp->next;
    }
}

/* Driver code*/
int main()
{
    /* Start with the empty list */
    struct Node* head = NULL;

    push(&head, 20);
    push(&head, 4);
    push(&head, 15);
    push(&head, 85);

    printf("Given linked list\n");
    printList(head);
    reverse(&head);
    printf("\nReversed linked list \n");
    printList(head);
    getchar();
}

Java

// Java program for reversing the linked list

class LinkedList {

    static Node head;

    static class Node {

        int data;
        Node next;

        Node(int d)
        {
            data = d;
            next = null;
        }
    }

    /* Function to reverse the linked list */
    Node reverse(Node node)
    {
        Node prev = null;
        Node current = node;
        Node next = null;
        while (current != null) {
            next = current.next;
            current.next = prev;
            prev = current;
            current = next;
        }
        node = prev;
        return node;
    }

    // prints content of double linked list
    void printList(Node node)
    {
        while (node != null) {
            System.out.print(node.data + " ");
            node = node.next;
        }
    }

    // Driver Code
    public static void main(String[] args)
    {
        LinkedList list = new LinkedList();
        list.head = new Node(85);
        list.head.next = new Node(15);
        list.head.next.next = new Node(4);
        list.head.next.next.next = new Node(20);

        System.out.println("Given linked list");
        list.printList(head);
        head = list.reverse(head);
        System.out.println("");
        System.out.println("Reversed linked list ");
        list.printList(head);
    }
}

// This code has been contributed by Mayank Jaiswal

Python

# Python program to reverse a linked list
# Time Complexity : O(n)
# Space Complexity : O(1)

# Node class
class Node:

    # Constructor to initialize the node object
    def __init__(self, data):
        self.data = data
        self.next = None


class LinkedList:

    # Function to initialize head
    def __init__(self):
        self.head = None

    # Function to reverse the linked list
    def reverse(self):
        prev = None
        current = self.head
        while(current is not None):
            next = current.next
            current.next = prev
            prev = current
            current = next
        self.head = prev

    # Function to insert a new node at the beginning
    def push(self, new_data):
        new_node = Node(new_data)
        new_node.next = self.head
        self.head = new_node

    # Utility function to print the LinkedList
    def printList(self):
        temp = self.head
        while(temp):
            print(temp.data, end=" ")
            temp = temp.next


# Driver code
llist = LinkedList()
llist.push(20)
llist.push(4)
llist.push(15)
llist.push(85)

print ("Given linked list")
llist.printList()
llist.reverse()
print ("\nReversed linked list")
llist.printList()

# This code is contributed by Nikhil Kumar Singh(nickzuck_007)

C#

// C# program for reversing the linked list
using System;

class GFG {

    // Driver Code
    static void Main(string[] args)
    {
        LinkedList list = new LinkedList();
        list.AddNode(new LinkedList.Node(85));
        list.AddNode(new LinkedList.Node(15));
        list.AddNode(new LinkedList.Node(4));
        list.AddNode(new LinkedList.Node(20));

        // List before reversal
        Console.WriteLine("Given linked list ");
        list.PrintList();

        // Reverse the list
        list.ReverseList();

        // List after reversal
        Console.WriteLine("Reversed linked list ");
        list.PrintList();
    }
}

class LinkedList {
    Node head;

    public class Node {
        public int data;
        public Node next;

        public Node(int d)
        {
            data = d;
            next = null;
        }
    }

    // function to add a new node at
    // the end of the list
    public void AddNode(Node node)
    {
        if (head == null)
            head = node;
        else {
            Node temp = head;
            while (temp.next != null) {
                temp = temp.next;
            }
            temp.next = node;
        }
    }

    // function to reverse the list
    public void ReverseList()
    {
        Node prev = null, current = head, next = null;
        while (current != null) {
            next = current.next;
            current.next = prev;
            prev = current;
            current = next;
        }
        head = prev;
    }

    // function to print the list data
    public void PrintList()
    {
        Node current = head;
        while (current != null) {
            Console.Write(current.data + " ");
            current = current.next;
        }
        Console.WriteLine();
    }
}

// This code is contributed by Mayank Sharma

JavaScript

<script>

// JavaScript program for reversing the linked list

var head;

     class Node {
        constructor(val) {
            this.data = val;
            this.next = null;
        }
    }

    /* Function to reverse the linked list */
    function reverse(node) {
    var prev = null;
    var current = node;
    var next = null;
        while (current != null) {
            next = current.next;
            current.next = prev;
            prev = current;
            current = next;
        }
        node = prev;
        return node;
    }

    // prints content of double linked list
    function printList(node) {
        while (node != null) {
            document.write(node.data + " ");
            node = node.next;
        }
    }

    // Driver Code
    
        head = new Node(85);
        head.next = new Node(15);
        head.next.next = new Node(4);
        head.next.next.next = new Node(20);

        document.write("Given linked list<br/>");
        printList(head);
        head = reverse(head);
        document.write("<br/>");
        document.write("Reversed linked list<br/> ");
        printList(head);

// This code is contributed by todaysgaurav

</script>

Output

Given linked list
85 15 4 20 
Reversed linked list 
20 4 15 85 

Time Complexity: O(N), Traversing over the linked list of size N. 
Auxiliary Space: O(1)

Reverse a linked list using Recursion:

The idea is to reach the last node of the linked list using recursion then start reversing the linked list.

Follow the steps below to solve the problem:

• Divide the list in two parts – first node and rest of the linked list.
• Call reverse for the rest of the linked list.
• Link the rest linked list to first.
• Fix head pointer to NULL

Below is the implementation of above approach:

C++

// Recursive C++ program to reverse
// a linked list
#include <bits/stdc++.h>
using namespace std;

/* Link list node */
struct Node {
    int data;
    struct Node* next;
    Node(int data)
    {
        this->data = data;
        next = NULL;
    }
};

struct LinkedList {
    Node* head;
    LinkedList() { head = NULL; }

    Node* reverse(Node* head)
    {
        if (head == NULL || head->next == NULL)
            return head;

        /* reverse the rest list and put
          the first element at the end */
        Node* rest = reverse(head->next);
        head->next->next = head;

        /* tricky step -- see the diagram */
        head->next = NULL;

        /* fix the head pointer */
        return rest;
    }

    /* Function to print linked list */
    void print()
    {
        struct Node* temp = head;
        while (temp != NULL) {
            cout << temp->data << " ";
            temp = temp->next;
        }
    }

    void push(int data)
    {
        Node* temp = new Node(data);
        temp->next = head;
        head = temp;
    }
};

/* Driver program to test above function*/
int main()
{
    /* Start with the empty list */
    LinkedList ll;
    ll.push(20);
    ll.push(4);
    ll.push(15);
    ll.push(85);

    cout << "Given linked list\n";
    ll.print();

    ll.head = ll.reverse(ll.head);

    cout << "\nReversed linked list \n";
    ll.print();
    return 0;
}

Java

// Recursive Java program to reverse
// a linked list

import java.io.*;

class recursion {
    static Node head; // head of list

    static class Node {
        int data;
        Node next;
        Node(int d)
        {
            data = d;
            next = null;
        }
    }

    static Node reverse(Node head)
    {
        if (head == null || head.next == null)
            return head;

        /* reverse the rest list and put
        the first element at the end */
        Node rest = reverse(head.next);
        head.next.next = head;

        /* tricky step -- see the diagram */
        head.next = null;

        /* fix the head pointer */
        return rest;
    }

    /* Function to print linked list */
    static void print()
    {
        Node temp = head;
        while (temp != null) {
            System.out.print(temp.data + " ");
            temp = temp.next;
        }
        System.out.println();
    }

    static void push(int data)
    {
        Node temp = new Node(data);
        temp.next = head;
        head = temp;
    }

    /* Driver program to test above function*/
    public static void main(String args[])
    {
        /* Start with the empty list */

        push(20);
        push(4);
        push(15);
        push(85);

        System.out.println("Given linked list");
        print();

        head = reverse(head);

        System.out.println("Reversed linked list");
        print();
    }
}

// This code is contributed by Prakhar Agarwal

Python

"""Python3 program to reverse linked list
using recursive method"""

# Linked List Node


class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

# Create and Handle list operations


class LinkedList:
    def __init__(self):
        self.head = None  # Head of list

    # Method to reverse the list
    def reverse(self, head):

        # If head is empty or has reached the list end
        if head is None or head.next is None:
            return head

        # Reverse the rest list
        rest = self.reverse(head.next)

        # Put first element at the end
        head.next.next = head
        head.next = None

        # Fix the header pointer
        return rest

    # Returns the linked list in display format
    def __str__(self):
        linkedListStr = ""
        temp = self.head
        while temp:
            linkedListStr = (linkedListStr +
                             str(temp.data) + " ")
            temp = temp.next
        return linkedListStr

    # Pushes new data to the head of the list
    def push(self, data):
        temp = Node(data)
        temp.next = self.head
        self.head = temp


# Driver code
linkedList = LinkedList()
linkedList.push(20)
linkedList.push(4)
linkedList.push(15)
linkedList.push(85)

print("Given linked list")
print(linkedList)

linkedList.head = linkedList.reverse(linkedList.head)

print("Reversed linked list")
print(linkedList)

# This code is contributed by Debidutta Rath

C#

// Recursive C# program to
// reverse a linked list
using System;
class recursion {

    // Head of list
    static Node head;

    public class Node {
        public int data;
        public Node next;
        public Node(int d)
        {
            data = d;
            next = null;
        }
    }

    static Node reverse(Node head)
    {
        if (head == null || head.next == null)
            return head;

        // Reverse the rest list and put
        // the first element at the end
        Node rest = reverse(head.next);
        head.next.next = head;

        // Tricky step --
        // see the diagram
        head.next = null;

        // Fix the head pointer
        return rest;
    }

    // Function to print linked list
    static void print()
    {
        Node temp = head;
        while (temp != null) {
            Console.Write(temp.data + " ");
            temp = temp.next;
        }
        Console.WriteLine();
    }

    static void push(int data)
    {
        Node temp = new Node(data);
        temp.next = head;
        head = temp;
    }

    // Driver code
    public static void Main(String[] args)
    {
        // Start with the
        // empty list
        push(20);
        push(4);
        push(15);
        push(85);

        Console.WriteLine("Given linked list");
        print();
        head = reverse(head);
        Console.WriteLine("Reversed linked list");
        print();
    }
}

// This code is contributed by gauravrajput1

JavaScript

<script>
// Recursive javascript program to reverse
// a linked list

    var head; // head of list
     class Node {
        constructor(val) {
            this.data = val;
            this.next = null;
        }
    }

    function reverse(head) {
        if (head == null || head.next == null)
            return head;

        /*
         * reverse the rest list and put the first element at the end
         */
        var rest = reverse(head.next);
        head.next.next = head;

        /* tricky step -- see the diagram */
        head.next = null;

        /* fix the head pointer */
        return rest;
    }

    /* Function to print linked list */
    function print() {
    var temp = head;
        while (temp != null) {
            document.write(temp.data + " ");
            temp = temp.next;
        }
        document.write();
    }

    function push(data) {
    var temp = new Node(data);
        temp.next = head;
        head = temp;
    }

    /* Driver program to test above function */
    
        /* Start with the empty list */

        push(20);
        push(4);
        push(15);
        push(85);

        document.write("Given linked list<br/>");
        print();

        head = reverse(head);

        document.write("<br/>Reversed Linked list<br/>");
        print();

// This code is contributed by Rajput-Ji
</script>

Output

Given linked list
85 15 4 20 
Reversed linked list 
20 4 15 85 

Time Complexity: O(N), Visiting over every node one time
Auxiliary Space: O(N), Function call stack space

Reverse a linked list by Tail Recursive Method:

The idea is to maintain three pointers previous, current and next, recursively visit every node and make links using these three pointers.

Follow the steps below to solve the problem:

• First update next with next node of current i.e. next = current->next
• Now make a reverse link from current node to previous node i.e. curr->next = prev
• If the visited node is the last node then just make a reverse link from the current node to previous node and update head. 

Below is the implementation of the above approach:

C++

// A simple and tail recursive C++ program to reverse
// a linked list
#include <bits/stdc++.h>
using namespace std;

struct Node {
    int data;
    struct Node* next;
      Node(int x) {
        data = x;
          next = NULL;
    }
};

void reverseUtil(Node* curr, Node* prev, Node** head);

// This function mainly calls reverseUtil()
// with prev as NULL
void reverse(Node** head)
{
    if (!head)
        return;
    reverseUtil(*head, NULL, head);
}

// A simple and tail-recursive function to reverse
// a linked list.  prev is passed as NULL initially.
void reverseUtil(Node* curr, Node* prev, Node** head)
{
    /* If last node mark it head*/
    if (!curr->next) {
        *head = curr;
        /* Update next to prev node */
        curr->next = prev;
        return;
    }
    /* Save curr->next node for recursive call */
    Node* next = curr->next;
    /* and update next ..*/
    curr->next = prev;
    reverseUtil(next, curr, head);
}

// A utility function to print a linked list
void printlist(Node* head)
{
    while (head != NULL) {
        cout << head->data << " ";
        head = head->next;
    }
    cout << endl;
}

// Driver code
int main()
{
    Node* head1 = new Node(1);
    head1->next = new Node(2);
    head1->next->next = new Node(3);
    head1->next->next->next = new Node(4);
    head1->next->next->next->next = new Node(5);
    head1->next->next->next->next->next = new Node(6);
    head1->next->next->next->next->next->next = new Node(7);
    head1->next->next->next->next->next->next->next
        = new Node(8);
    cout << "Given linked list\n";
    printlist(head1);
    reverse(&head1);
    cout << "Reversed linked list\n";
    printlist(head1);
    return 0;
}

C

// A simple and tail recursive C program to reverse a linked
// list
#include <stdio.h>
#include <stdlib.h>

typedef struct Node {
    int data;
    struct Node* next;
} Node;

void reverseUtil(Node* curr, Node* prev, Node** head);

// This function mainly calls reverseUtil()
// with prev as NULL
void reverse(Node** head)
{
    if (!head)
        return;
    reverseUtil(*head, NULL, head);
}

// A simple and tail-recursive function to reverse
// a linked list.  prev is passed as NULL initially.
void reverseUtil(Node* curr, Node* prev, Node** head)
{
    /* If last node mark it head*/
    if (!curr->next) {
        *head = curr;
        /* Update next to prev node */
        curr->next = prev;
        return;
    }

    /* Save curr->next node for recursive call */
    Node* next = curr->next;
    /* and update next ..*/
    curr->next = prev;
    reverseUtil(next, curr, head);
}

// A utility function to create a new node
Node* newNode(int key)
{
    Node* temp = (Node*)malloc(sizeof(Node));
    temp->data = key;
    temp->next = NULL;
    return temp;
}

// A utility function to print a linked list
void printlist(Node* head)
{
    while (head != NULL) {
        printf("%d ", head->data);
        head = head->next;
    }
    printf("\n");
}

// Driver code
int main()
{
    Node* head1 = newNode(1);
    head1->next = newNode(2);
    head1->next->next = newNode(3);
    head1->next->next->next = newNode(4);
    head1->next->next->next->next = newNode(5);
    head1->next->next->next->next->next = newNode(6);
    head1->next->next->next->next->next->next = newNode(7);
    head1->next->next->next->next->next->next->next
        = newNode(8);
    printf("Given linked list\n");
    printlist(head1);
    reverse(&head1);
    printf("Reversed linked list\n");
    printlist(head1);
    return 0;
}

// This code is contributed by Aditya Kumar (adityakumar129)

Java

// Java program for reversing the Linked list

class LinkedList {

    static Node head;
    static class Node {
        int data;
        Node next;
        Node(int d)
        {
            data = d;
            next = null;
        }
    }
    // A simple and tail recursive function to reverse
    // a linked list.  prev is passed as NULL initially.
    Node reverseUtil(Node curr, Node prev)
    {
        /*If head is initially null OR list is empty*/
        if (head == null)
            return head;
        /* If last node mark it head*/
        if (curr.next == null) {
            head = curr;
            /* Update next to prev node */
            curr.next = prev;
            return head;
        }
        /* Save curr->next node for recursive call */
        Node next1 = curr.next;
        /* and update next ..*/
        curr.next = prev;
        reverseUtil(next1, curr);
        return head;
    }

    // prints content of double linked list
    void printList(Node node)
    {
        while (node != null) {
            System.out.print(node.data + " ");
            node = node.next;
        }
    }

    // Driver Code
    public static void main(String[] args)
    {
        LinkedList list = new LinkedList();
        list.head = new Node(1);
        list.head.next = new Node(2);
        list.head.next.next = new Node(3);
        list.head.next.next.next = new Node(4);
        list.head.next.next.next.next = new Node(5);
        list.head.next.next.next.next.next = new Node(6);
        list.head.next.next.next.next.next.next
            = new Node(7);
        list.head.next.next.next.next.next.next.next
            = new Node(8);

        System.out.println("Given linked list ");
        list.printList(head);
        Node res = list.reverseUtil(head, null);
        System.out.println("\nReversed linked list ");
        list.printList(res);
    }
}

// This code is contributed by Aditya Kumar (adityakumar129)

Python

# Simple and tail recursive Python program to
# reverse a linked list

# Node class


class Node:

    # Constructor to initialize the node object
    def __init__(self, data):
        self.data = data
        self.next = None


class LinkedList:

    # Function to initialize head
    def __init__(self):
        self.head = None

    def reverseUtil(self, curr, prev):

        # If last node mark it head
        if curr.next is None:
            self.head = curr

            # Update next to prev node
            curr.next = prev
            return

        # Save curr.next node for recursive call
        next = curr.next

        # And update next
        curr.next = prev

        self.reverseUtil(next, curr)

    # This function mainly calls reverseUtil()
    # with previous as None

    def reverse(self):
        if self.head is None:
            return
        self.reverseUtil(self.head, None)

    # Function to insert a new node at the beginning

    def push(self, new_data):
        new_node = Node(new_data)
        new_node.next = self.head
        self.head = new_node

    # Utility function to print the linked LinkedList
    def printList(self):
        temp = self.head
        while(temp):
            print (temp.data, end=" ")
            temp = temp.next


# Driver code
llist = LinkedList()
llist.push(8)
llist.push(7)
llist.push(6)
llist.push(5)
llist.push(4)
llist.push(3)
llist.push(2)
llist.push(1)

print ("Given linked list")
llist.printList()

llist.reverse()

print ("\nReversed linked list")
llist.printList()

# This code is contributed by Nikhil Kumar Singh(nickzuck_007)

C#

// C# program for reversing the Linked list
using System;

public class LinkedList {
    Node head;
    public class Node {

        public int data;
        public Node next;

        public Node(int d)
        {
            data = d;
            next = null;
        }
    }

    // A simple and tail-recursive function to reverse
    // a linked list. prev is passed as NULL initially.
    Node reverseUtil(Node curr, Node prev)
    {

        /* If last node mark it head*/
        if (curr.next == null) {
            head = curr;

            /* Update next to prev node */
            curr.next = prev;

            return head;
        }

        /* Save curr->next node for recursive call */
        Node next1 = curr.next;

        /* and update next ..*/
        curr.next = prev;

        reverseUtil(next1, curr);
        return head;
    }

    // prints content of double linked list
    void printList(Node node)
    {
        while (node != null) {
            Console.Write(node.data + " ");
            node = node.next;
        }
    }

    // Driver code
    public static void Main(String[] args)
    {
        LinkedList list = new LinkedList();
        list.head = new Node(1);
        list.head.next = new Node(2);
        list.head.next.next = new Node(3);
        list.head.next.next.next = new Node(4);
        list.head.next.next.next.next = new Node(5);
        list.head.next.next.next.next.next = new Node(6);
        list.head.next.next.next.next.next.next
            = new Node(7);
        list.head.next.next.next.next.next.next.next
            = new Node(8);

        Console.WriteLine("Given linked list ");
        list.printList(list.head);
        Node res = list.reverseUtil(list.head, null);
        Console.WriteLine("\nReversed linked list ");
        list.printList(res);
    }
}

// This code contributed by Rajput-Ji

JavaScript

<script>
// javascript program for reversing the Linked list

    var head;
 class Node {
        constructor(d) {
            this.data = d;
            this.next = null;
        }
    }

    // A simple and tail recursive function to reverse
    // a linked list. prev is passed as NULL initially.
    function reverseUtil(curr,  prev) 
    {
    
        /* If head is initially null OR list is empty */
        if (head == null)
            return head;
            
        /* If last node mark it head */
        if (curr.next == null) {
            head = curr;

            /* Update next to prev node */
            curr.next = prev;
            return head;
        }

        /* Save curr->next node for recursive call */
        var next1 = curr.next;

        /* and update next .. */
        curr.next = prev;

        reverseUtil(next1, curr);
        return head;
    }

    // prints content of var linked list
    function printList(node) {
        while (node != null) {
            document.write(node.data + " ");
            node = node.next;
        }
    }

    // Driver Code
        var head = new Node(1);
        head.next = new Node(2);
        head.next.next = new Node(3);
        head.next.next.next = new Node(4);
        head.next.next.next.next = new Node(5);
        head.next.next.next.next.next = new Node(6);
        head.next.next.next.next.next.next = new Node(7);
        head.next.next.next.next.next.next.next = new Node(8);

        document.write("Original Linked list<br/> ");
        printList(head);
        var res = reverseUtil(head, null);
        document.write("<br/>");
        document.write("Reversed linked list <br/>");
        printList(res);

// This code is contributed by Rajput-Ji 
</script>

Output

Given linked list
1 2 3 4 5 6 7 8 
Reversed linked list
8 7 6 5 4 3 2 1 

Time Complexity: O(N), Visiting every node of the linked list of size N.
Auxiliary Space: O(N), Function call stack space. If tail call elimination happens, then we can say that the auxiliary space is O(1)

Reverse a linked list using Stack:

The idea is to store the all the nodes in the stack then make a reverse linked list.

Follow the steps below to solve the problem:

• Store the nodes(values and address) in the stack until all the values are entered.
• Once all entries are done, Update the Head pointer to the last location(i.e the last value).
• Start popping the nodes(value and address) and store them in the same order until the stack is empty.
• Update the next pointer of last Node in the stack by NULL.

Below is the implementation of the above approach:

C++

// C++ program for above approach
#include <bits/stdc++.h>
#include <iostream>
using namespace std;

// Create a class Node to enter values and address in the
// list
class Node {
public:
    int data;
    Node* next;
      Node(int x) {
        data = x;
         next = NULL;
    }
};

// Function to reverse the linked list
void reverseLL(Node** head)
{
    // Create a stack "s" of Node type
    stack<Node*> s;
    Node* temp = *head;
    while (temp->next != NULL) {
        // Push all the nodes in to stack
        s.push(temp);
        temp = temp->next;
    }
    *head = temp;
    while (!s.empty()) {
        // Store the top value of stack in list
        temp->next = s.top();
        // Pop the value from stack
        s.pop();
        // update the next pointer in the list
        temp = temp->next;
    }
    temp->next = NULL;
}

// Function to Display the elements in List
void printlist(Node* temp)
{
    while (temp != NULL) {
        cout << temp->data << " ";
        temp = temp->next;
    }
}

// Program to insert back of the linked list
void insert_back(Node** head, int value)
{

    // we have used insertion at back method to enter values
    // in the list.(eg: head->1->2->3->4->Null)
    Node* temp = new Node(value);
    temp->next = NULL;

    // If *head equals to NULL
    if (*head == NULL) {
        *head = temp;
        return;
    }
    else {
        Node* last_node = *head;
        while (last_node->next != NULL)
            last_node = last_node->next;
        last_node->next = temp;
        return;
    }
}

// Driver Code
int main()
{
    Node* head = NULL;
    insert_back(&head, 1);
    insert_back(&head, 2);
    insert_back(&head, 3);
    insert_back(&head, 4);
    cout << "Given linked list\n";
    printlist(head);
    reverseLL(&head);
    cout << "\nReversed linked list\n";
    printlist(head);
    return 0;
}

// This code is contributed by Aditya Kumar (adityakumar129)

Java

// Java program for above approach
import java.util.*;

class GFG {
    // Create a class Node to enter values and address in
    // the list
    static class Node {
        int data;
        Node next;
          Node(int x) {
            data = x;
              next = null;
        }
    };
    static Node head = null;
    // Function to reverse the linked list
    static void reverseLL()
    {

        // Create a stack "s" of Node type
        Stack<Node> s = new Stack<>();
        Node temp = head;
        while (temp.next != null) {
            // Push all the nodes in to stack
            s.add(temp);
            temp = temp.next;
        }
        head = temp;
        while (!s.isEmpty()) {
            // Store the top value of stack in list
            temp.next = s.peek();
            // Pop the value from stack
            s.pop();
            // update the next pointer in the list
            temp = temp.next;
        }
        temp.next = null;
    }

    // Function to Display the elements in List
    static void printlist(Node temp)
    {
        while (temp != null) {
            System.out.print(temp.data + " ");
            temp = temp.next;
        }
    }

    // Program to insert back of the linked list
    static void insert_back(int value)
    {
        // we have used insertion at back method to enter
        // values in the list.(eg: head.1.2.3.4.Null)
        Node temp = new Node(value);
        temp.next = null;
        // If *head equals to null
        if (head == null) {
            head = temp;
            return;
        }
        else {
            Node last_node = head;
            while (last_node.next != null)
                last_node = last_node.next;
            last_node.next = temp;
            return;
        }
    }

    // Driver Code
    public static void main(String[] args)
    {
        insert_back(1);
        insert_back(2);
        insert_back(3);
        insert_back(4);
        System.out.print("Given linked list\n");
        printlist(head);
        reverseLL();
        System.out.print("\nReversed linked list\n");
        printlist(head);
    }
}

// This code is contributed by Aditya Kumar (adityakumar129)

Python

# Python code for the above approach

# Definition for singly-linked list.


class ListNode:
    def __init__(self, val = 0, next=None):
        self.val = val
        self.next = next


class Solution:

    # Program to reverse the linked list
    # using stack
    def reverseLLUsingStack(self, head):

        # Initialise the variables
        stack, temp = [], head

        while temp:
            stack.append(temp)
            temp = temp.next

        head = temp = stack.pop()

        # Until stack is not
        # empty
        while len(stack) > 0:
            temp.next = stack.pop()
            temp = temp.next

        temp.next = None
        return head


# Driver Code
if __name__ == "__main__":
    head = ListNode(1, ListNode(2, ListNode(3, ListNode(4))))
    print("Given linked list")
    temp = head
    while temp:
        print(temp.val, end=' ')
        temp = temp.next
    obj = Solution()
    print("\nReversed linked list")
    head = obj.reverseLLUsingStack(head)
    while head:
        print(head.val, end=' ')
        head = head.next

C#

// C# program for above approach
using System;
using System.Collections.Generic;

class GFG {

    // Create a class Node to enter
    // values and address in the list
    public class Node {
        public int data;
        public Node next;
          public Node(int x) {
            data = x;
        }
    };
    static Node head = null;

    // Function to reverse the
    // linked list
    static void reverseLL()
    {

        // Create a stack "s"
        // of Node type
        Stack<Node> s = new Stack<Node>();
        Node temp = head;

        while (temp.next != null) {

            // Push all the nodes
            // in to stack
            s.Push(temp);
            temp = temp.next;
        }
        head = temp;

        while (s.Count != 0) {

            // Store the top value of
            // stack in list
            temp.next = s.Peek();

            // Pop the value from stack
            s.Pop();

            // Update the next pointer in the
            // in the list
            temp = temp.next;
        }
        temp.next = null;
    }

    // Function to Display
    // the elements in List
    static void printlist(Node temp)
    {
        while (temp != null) {
            Console.Write(temp.data + " ");
            temp = temp.next;
        }
    }

    // Function to insert back of the
    // linked list
    static void insert_back(int val)
    {

        // We have used insertion at back method
        // to enter values in the list.(eg:
        // head.1.2.3.4.Null)
        Node temp = new Node(val);
        temp.next = null;

        // If *head equals to null
        if (head == null) {
            head = temp;
            return;
        }
        else {
            Node last_node = head;

            while (last_node.next != null) {
                last_node = last_node.next;
            }
            last_node.next = temp;
            return;
        }
    }

    // Driver Code
    public static void Main(String[] args)
    {
        insert_back(1);
        insert_back(2);
        insert_back(3);
        insert_back(4);
        Console.Write("Given linked list\n");

        printlist(head);
        reverseLL();

        Console.Write("\nReversed linked list\n");
        printlist(head);
    }
}

// This code is contributed by gauravrajput1

JavaScript

<script>
// javascript program for above approach

// Create a class Node to enter 
// values and address in the list
 class Node 
{
     constructor(x){
    this.data = x;
    this.next = null;
}
}
var head = null;
// Function to reverse the 
// linked list
function reverseLL()
{   
    
    // Create a stack "s" 
    // of Node type
    var s = []; 
    var temp = head;
    while (temp.next != null) 
    {
        
        // Push all the nodes 
        // in to stack
        s.push(temp); 
        temp = temp.next;
    }
    head = temp;
  
    while (s.length!=0) 
    {
        
        // Store the top value of
        // stack in list
        temp.next = s.pop(); 
      
      
      
        // update the next pointer in the
        // in the list
        temp = temp.next; 
    }
    temp.next = null;
}

// Function to Display 
// the elements in List
function printlist(temp) 
{
    while (temp != null) 
    {
        document.write(temp.data+ " ");
        temp = temp.next;
    }
}

// Program to insert back of the 
// linked list
function insert_back(  value)
{ 

    // we have used insertion at back method
    // to enter values in the list.(eg:
    // head.1.2.3.4.Null)
    var temp = new Node(value);
    temp.next = null;
    
    // If *head equals to null
    if (head == null) 
    {
      head = temp;
      return;
    }
    else 
    {
      var last_node = head;
      while (last_node.next != null) 
      {
        last_node = last_node.next;
      }
      last_node.next = temp;
      return;
    }
}

// Driver Code

        insert_back(1);
        insert_back(2);
        insert_back(3);
        insert_back(4);
        document.write("Given linked list\n");
        printlist(head);
        reverseLL();
        document.write("<br/>Reversed linked list\n");
        printlist(head);

 // This code is contributed by umadevi9616 
</script>

Output

Given linked list
1 2 3 4 
Reversed linked list
4 3 2 1 

Time Complexity: O(N), Visiting every node of the linked list of size N.
Auxiliary Space: O(N), Space is used to store the nodes in the stack.