Reverse a linked list

Given 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 

Iterative Method

1. Initialize three pointers prev as NULL, curr as head and next as NULL.
2. Iterate through the linked list. In loop, do following. 
   // Before changing next of current, 
   // store next node 
   next = curr->next
   // Now change next of current 
   // This is where actual reversing happens 
   curr->next = prev 
   // Move prev and curr one step forward 
   prev = curr 
   curr = next

Below is the implementation of the above approach: 

C++

// Iterative C++ program to reverse a linked list
#include <iostream>
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 linked LinkedList
    def printList(self):
        temp = self.head
        while(temp):
            print temp.data,
            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) 
Auxiliary Space: O(1)

Recursive Method:

   1) Divide the list in two parts - first node and 
      rest of the linked list.
   2) Call reverse for the rest of the linked list.
   3) Link rest to first.
   4) Fix head pointer

C++

// Recursive C++ program to reverse
// a linked list
#include <iostream>
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
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

Python3

"""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)
Auxiliary Space: O(n)

A Simpler and Tail Recursive Method

Below is the implementation of this method.  

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;
};

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 = new Node;
    temp->data = key;
    temp->next = NULL;
    return temp;
}

// 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 = 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);
    cout << "Given linked list\n";
    printlist(head1);
    reverse(&head1);
    cout << "\nReversed linked list\n";
    printlist(head1);
    return 0;
}

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

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("\nReversed 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("Original Linked list ");
        list.printList(head);
        Node res = list.reverseUtil(head, null);
        System.out.println("");
        System.out.println("");
        System.out.println("Reversed 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,
            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 "\nReverse 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("Original Linked list ");
        list.printList(list.head);
        Node res = list.reverseUtil(list.head, null);
        Console.WriteLine("");
        Console.WriteLine("");
        Console.WriteLine("Reversed 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("<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)

As we do certain operation for every node of the linked list.

Auxiliary Space: O(1)

As constant extra space is used.

Head Recursive Method:

C++

// Head recursive C++ program to reverse a linked list
#include <iostream>
using namespace std;

// Linked list node
class Node {
public:
    int data;
    Node* next;

    // constructor: automatically assigns the value to the
    // data and next pointer to NULL
    Node(){};
    Node(int val)
        : data(val)
        , next(NULL){};
};

void reverseUtil(Node* curr, Node* prev, Node** headref)
{
    // Base Case - If curr is last node
    if (curr->next == NULL) {
        // Update head of the linked list
        *headref = curr;
        // Update next to prev node
        curr->next = prev;
        return;
    }
    // Recursive Call for next node
    reverseUtil(curr->next, curr, headref);
    // Update next to prev node
    curr->next = prev;
}

void reverse(Node** headref)
{
    // If linked list is empty or contains single node
    if (*headref == NULL || (*headref)->next == NULL)
        return;
    // Call reverseUtil() with prev as NULL
    reverseUtil(*headref, NULL, headref);
}

// Function to insert a node at the end of linked list
void push(Node** headref, int x)
{
    Node* newptr = new Node(x);
    if (*headref == NULL) {
        *headref = newptr;
    }
    else {
        Node* temp = *headref;
        while (temp->next != NULL) {
            temp = temp->next;
        }
        temp->next = newptr;
    }
}

// Functio to print the linked list
void print(Node* headref)
{
    while (headref != NULL) {
        cout << headref->data << " ";
        headref = headref->next;
    }
    cout << "\n";
}

int main()
{
    Node* head = NULL;

    // head->1->2->3->4->5->6->NULL
    push(&head, 1);
    push(&head, 2);
    push(&head, 3);
    push(&head, 4);
    push(&head, 5);
    push(&head, 6);

    cout << "Given Linked List\n";
    print(head);

    reverse(&head);
    cout << "Reversed Linked List\n";
    print(head);
    return 0;
}

//This code is contributed by Anisha Wagh

Output:

Given Linked List
1 2 3 4 5 6

Reversed Linked List
6 5 4 3 2 1

Time Complexity : O(N)

Using Stack:

Algorithm – 

• 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;
};

// 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 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();
    temp->data = 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;
    };
    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 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();
        temp.data = 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)

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;
};
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 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();
    temp.data = 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);
    Console.Write("Given linked list\n");

    printlist(head);
    reverseLL();

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

// This code is contributed by gauravrajput1

Python3

# 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, ListNode(5)))))
    obj = Solution()
    head = obj.reverseLLUsingStack(head)
    while head:
        print(head.val, end=' ')
        head = head.next

Javascript

<script>
// javascript program for above approach

// Create a class Node to enter
// values and address in the list
 class Node
{
     constructor(){
    this.data = 0;
    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();
    temp.data = 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 

Thanks to Gaurav Ahirwar for suggesting this solution. 

Time Complexity: O(N)

As we do certain operation for every node of the linked list.

Auxiliary Space: O(N)

Space is used to store the nodes in the stack.

Using array:

Algorithm :- 

1. Create a linked list.

2. Count the number of nodes present in the Linked List

3. Initialize an array with the size of the count.

4. Store the elements of the Linked list in array 

5. Print the array from the last index to the first.

C++

#include <bits/stdc++.h>
using namespace std;

typedef struct node {
    int val;
    struct node* next;
} node;

node* head = NULL;

// Function to return the No of nodes present in the linked list
int count(node* head)
{
    node* p = head;
    int k = 1;
    while (p != NULL) {
        p = p->next;
        k++;
    }
    return k;
}

node* ll_reverse(node* head) // to reverse the linked list
{
    node* p = head;
    long int i = count(head), j = 1;
    long int arr[i];
    while (i && p != NULL) {
        arr[j++] = p->val;
        p = p->next;
        i--;
    }
    j--;
    while (j) // loop will break as soon as j=0
        cout << arr[j--] << " ";
    return head;
}

// Function to insert node at the end of linked list
node* insert_end(node* head, int data)
{
    node *q = head, *p = (node*)malloc(sizeof(node));
    p->val = data;
    while (q->next != NULL)
        q = q->next;
    q->next = p;
    p->next = NULL;
    return head;
}

node* create_ll(node* head, int data) // create ll
{
    node* p = (node*)malloc(sizeof(node));
    p->val = data;
    if (head == NULL) {
        head = p;
        p->next = NULL;
        return head;
    }
    else {
        head = insert_end(head, data);
        return head;
    }
}

// Driver code

int main()
{
    int i = 5, j = 1;
    while (i--)
        head = create_ll(head, j++);
    head = ll_reverse(head);
    return 0;
}

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

C

#include<stdio.h>
#include<stdlib.h>

typedef struct node {
    int val;
    struct node* next;
} node;

node* head = NULL;

// Function to return the No of nodes present in the linked list
int count(node* head)
{
    node* p = head;
    int k = 1;
    while (p != NULL) {
        p = p->next;
        k++;
    }
    return k;
}

node* ll_reverse(node* head) // to reverse the linked list
{
    node* p = head;
    long int i = count(head), j = 1;
    int arr[i];
    while (i && p != NULL) {
        arr[j++] = p->val;
        p = p->next;
        i--;
    }
    j--;
    while (j) // loop will break as soon as j=0
        printf("%d ",arr[j--]);
    return head;
}

// Function to insert node at the end of linked list
node* insert_end(node* head, int data)
{
    node *q = head, *p = (node*)malloc(sizeof(node));
    p->val = data;
    while (q->next != NULL)
        q = q->next;
    q->next = p;
    p->next = NULL;
    return head;
}

node* create_ll(node* head, int data) // create ll
{
    node* p = (node*)malloc(sizeof(node));
    p->val = data;
    if (head == NULL) {
        head = p;
        p->next = NULL;
        return head;
    }
    else {
        head = insert_end(head, data);
        return head;
    }
}

// Driver code

int main()
{
    int i = 5, j = 1;
    while (i--)
        head = create_ll(head, j++);
    head = ll_reverse(head);
    return 0;
}

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

Java

// Java program of the above approach
class GFG {

    // Create a class Node to enter values and address in the list
    static class node {
        int val;
        node next;
    };

    static node head = null;
    // code to count the no. of nodes
    static int count(node head)
    {
        node p = head;
        int k = 1;
        while (p != null) {
            p = p.next;
            k++;
        }
        return k;
    }

    // to reverse the linked list
    static node ll_reverse(node head)
    {
        node p = head;
        int i = count(head), j = 1;
        int[] arr = new int[i];
        while (i != 0 && p != null) {
            arr[j++] = p.val;
            p = p.next;
            i--;
        }
        j--;
        while (j != 0) // loop will break as soon as j=0
            System.out.print(arr[j--] + " ");
        return head;
    }
    // code to insert at end of ll
    static node insert_end(node head, int data)
    {
        node q = head;
        node p = new node();
        p.val = data;
        p.next = null;
        while (q.next != null)
            q = q.next;
        q.next = p;
        p.next = null;
        return head;
    }

    // create ll
    static node create_ll(node head, int data)
    {
        node p = new node();
        p.next = null;
        p.val = data;
        if (head == null) {
            head = p;
            p.next = null;
            return head;
        }
        else {
            head = insert_end(head, data);
            return head;
        }
    }

    public static void main(String[] args)
    {
        int i = 5, j = 1;
        while (i != 0) {
            head = create_ll(head, j++);
            i--;
        }
        head = ll_reverse(head);
    }
}

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

C#

// C# program of the above approach
using System;

public class GFG {

    // Create a class Node to enter
    // values and address in the list
    public

 class node {

        public

 int val;
        public

 node next;
    };

    static node head = null;

    // code to count the no. of nodes
    static int count(node head)
    {
        node p = head;
        int k = 1;
        while (p != null) {
            p = p.next;
            k++;
        }
        return k;
    }

    // to reverse the linked list
    static node ll_reverse(node head)
    {
        node p = head;
        int i = count(head), j = 1;
        int[] arr = new int[i];

        while (i != 0 && p != null) {
            arr[j++] = p.val;
            p = p.next;
            i--;
        }
        j--;
        while (j != 0) // loop will break as soon as j=0
        {
            Console.Write(arr[j--] + " ");
        }

        return head;
    }
    // code to insert at end of ll
    static node insert_end(node head, int data)
    {
        node q = head;
        node p = new node();
        p.val = data;
        p.next = null;
        while (q.next != null) {
            q = q.next;
        }
        q.next = p;
        p.next = null;
        return head;
    }

    // create ll
    static node create_ll(node head, int data)
    {
        node p = new node();
        p.next = null;
        p.val = data;
        if (head == null) {
            head = p;
            p.next = null;
            return head;
        }
        else {
            head = insert_end(head, data);
            return head;
        }
    }

    public static void Main(String[] args)
    {
        int i = 5, j = 1;
        while (i != 0) {
            head = create_ll(head, j++);
            i--;
        }
        head = ll_reverse(head);
    }
}

// This code is contributed by umadevi9616

Javascript

<script>
// Javascript program of the above approach

// Create a class Node to enter values and address in the list
class node {
  constructor() {
    this.val = null;
    this.next = null;
  }
};

let head = null;
// code to count the no. of nodes
function count(head) {
  let p = head;
  let k = 1;
  while (p != null) {
    p = p.next;
    k++;
  }
  return k;
}

// to reverse the linked list
function ll_reverse(head) {
  let p = head;
  let i = count(head), j = 1;
  let arr = new Array(i);
  while (i != 0 && p != null) {
    arr[j++] = p.val;
    p = p.next;
    i--;
  }
  j--;
  while (j != 0) // loop will break as soon as j=0
    document.write(arr[j--] + " ");
  return head;
}
// code to insert at end of ll
function insert_end(head, data) {
  let q = head;
  let p = new node();
  p.val = data;
  p.next = null;
  while (q.next != null)
    q = q.next;
  q.next = p;
  p.next = null;
  return head;
}

// create ll
function create_ll(head, data) {
  let p = new node();
  p.next = null;
  p.val = data;
  if (head == null) {
    head = p;
    p.next = null;
    return head;
  }
  else {
    head = insert_end(head, data);
    return head;
  }
}

let i = 5, j = 1;
while (i != 0) {
  head = create_ll(head, j++);
  i--;
}
head = ll_reverse(head);

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

Input :  1->2->3->4->5
Output: 5->4->3->2->1

Time complexity: O(N) as we visit every node once.

Auxiliary Space: O(N) as extra space is used to store all the nodes in the array.

Recursively Reversing a linked list (A simple implementation)
Iteratively Reverse a linked list using only 2 pointers (An Interesting Method)

References:
http://cslibrary.stanford.edu/105/LinkedListProblems.pdf