Rearrange a given linked list in-place.

Given a singly linked list L₀ -> L₁ -> … -> L_n-1 -> L_n. Rearrange the nodes in the list so that the new formed list is : L₀ -> L_n -> L₁ -> L_n-1 -> L₂ -> L_n-2 … You are required to do this in place without altering the nodes’ values.

Examples:

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

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

Simple Solution

1) Initialize current node as head.
2) While next of current node is not null, do following
    a) Find the last node, remove it from the end and insert it as next
       of the current node.
    b) Move current to next of current

The time complexity of the above simple solution is O(n²) where n is the number of nodes in the linked list.

Better Solution:

Copy contents of the given linked list to a vector.
Rearrange the given vector by swapping nodes from both ends.
Copy the modified vector back to the linked list.

Implementation of this approach: https://ide.geeksforgeeks.org/1eGSEy

Efficient Solution:

1) Find the middle point using tortoise and hare method.
2) Split the linked list into two halves using found middle point in step 1.
3) Reverse the second half.
4) Do alternate merge of first and second halves.

Below is the implementation of this method.

C++

// C++ program to rearrange a linked list in-place
#include <bits/stdc++.h>
using namespace std;
 
// Linkedlist Node structure
struct Node {
    int data;
    struct Node* next;
};
 
// Function to create newNode in a linkedlist
Node* newNode(int key)
{
    Node* temp = new Node;
    temp->data = key;
    temp->next = NULL;
    return temp;
}
 
// Function to reverse the linked list
void reverselist(Node** head)
{
    // Initialize prev and current pointers
    Node *prev = NULL, *curr = *head, *next;
 
    while (curr) {
        next = curr->next;
        curr->next = prev;
        prev = curr;
        curr = next;
    }
    *head = prev;
}
 
// Function to print the linked list
void printlist(Node* head)
{
    while (head != NULL) {
        cout << head->data << " ";
        if (head->next)
            cout << "-> ";
        head = head->next;
    }
    cout << endl;
}
 
// Function to rearrange a linked list
void rearrange(Node** head)
{
    // 1) Find the middle point using tortoise and hare
    // method
    Node *slow = *head, *fast = slow->next;
    while (fast && fast->next) {
        slow = slow->next;
        fast = fast->next->next;
    }
    // 2) Split the linked list in two halves
    // head1, head of first half    1 -> 2
    // head2, head of second half   3 -> 4
    Node* head1 = *head;
    Node* head2 = slow->next;
    slow->next = NULL;
    // 3) Reverse the second half, i.e.,  4 -> 3
    reverselist(&head2);
    // 4) Merge alternate nodes
    *head = newNode(0); // Assign dummy Node
    // curr is the pointer to this dummy Node, which will
    // be used to form the new list
    Node* curr = *head;
    while (head1 || head2) {
        // First add the element from list
        if (head1) {
            curr->next = head1;
            curr = curr->next;
            head1 = head1->next;
        }
        // Then add the element from the second list
        if (head2) {
            curr->next = head2;
            curr = curr->next;
            head2 = head2->next;
        }
    }
    // Assign the head of the new list to head pointer
    *head = (*head)->next;
}
 
// Driver program
int main()
{
    Node* head = newNode(1);
    head->next = newNode(2);
    head->next->next = newNode(3);
    head->next->next->next = newNode(4);
    head->next->next->next->next = newNode(5);
 
    printlist(head); // Print original list
    rearrange(&head); // Modify the list
    printlist(head); // Print modified list
    return 0;
}
 
// This code is contributed by Aditya Kumar (adityakumar129)

Java

// Java program to rearrange link list in place
 
// Linked List Class
class LinkedList {
 
    static Node head; // head of the list
 
    /* Node Class */
    static class Node {
 
        int data;
        Node next;
 
        // Constructor to create a new node
        Node(int d)
        {
            data = d;
            next = null;
        }
    }
 
    void printlist(Node node)
    {
        if (node == null) {
            return;
        }
        while (node != null) {
            System.out.print(node.data + " -> ");
            node = node.next;
        }
    }
 
    Node reverselist(Node node)
    {
        Node prev = null, curr = node, next;
        while (curr != null) {
            next = curr.next;
            curr.next = prev;
            prev = curr;
            curr = next;
        }
        node = prev;
        return node;
    }
 
    void rearrange(Node node)
    {
 
        // 1) Find the middle point using tortoise and hare
        // method
        Node slow = node, fast = slow.next;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
 
        // 2) Split the linked list in two halves
        // node1, head of first half    1 -> 2 -> 3
        // node2, head of second half   4 -> 5
        Node node1 = node;
        Node node2 = slow.next;
        slow.next = null;
 
        // 3) Reverse the second half, i.e., 5 -> 4
        node2 = reverselist(node2);
 
        // 4) Merge alternate nodes
        node = new Node(0); // Assign dummy Node
 
        // curr is the pointer to this dummy Node, which
        // will be used to form the new list
        Node curr = node;
        while (node1 != null || node2 != null) {
 
            // First add the element from first list
            if (node1 != null) {
                curr.next = node1;
                curr = curr.next;
                node1 = node1.next;
            }
 
            // Then add the element from second list
            if (node2 != null) {
                curr.next = node2;
                curr = curr.next;
                node2 = node2.next;
            }
        }
 
        // Assign the head of the new list to head pointer
        node = node.next;
    }
 
    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.printlist(head); // print original list
        list.rearrange(head); // rearrange list as per ques
        System.out.println("");
        list.printlist(head); // print modified list
    }
}
 
// This code has been contributed by Mayank Jaiswal

Python3

# Python program to rearrange link list in place
 
# Node Class
class Node:
   
    # Constructor to create a new node
    def __init__(self, d):
        self.data = d
        self.next = None
         
def printlist(node):
    if(node == None):
        return
    while(node != None):
        print(node.data," -> ", end = "")
        node = node.next
 
def reverselist(node):
    prev = None
    curr = node
    next=None
    while (curr != None):
        next = curr.next
        curr.next = prev
        prev = curr
        curr = next
    node = prev
    return node
 
def rearrange(node):
   
    # 1) Find the middle point using tortoise and hare
    # method
    slow = node
    fast = slow.next
    while (fast != None and fast.next != None):
        slow = slow.next
        fast = fast.next.next
     
    # 2) Split the linked list in two halves
    # node1, head of first half    1 -> 2 -> 3
    # node2, head of second half   4 -> 5   
    node1 = node
    node2 = slow.next
    slow.next = None
     
    # 3) Reverse the second half, i.e., 5 -> 4
    node2 = reverselist(node2)
     
    # 4) Merge alternate nodes
    node = Node(0)  #Assign dummy Node
     
    # curr is the pointer to this dummy Node, which
    # will be used to form the new list
    curr = node
     
    while (node1 != None or node2 != None):
         
        # First add the element from first list
        if (node1 != None):
            curr.next = node1
            curr = curr.next
            node1 = node1.next
         
        # Then add the element from second list
        if(node2 != None):
            curr.next = node2
            curr = curr.next
            node2 = node2.next
     
    # Assign the head of the new list to head pointer
    node = node.next
 
head = None
head = Node(1)
head.next = Node(2)
head.next.next = Node(3)
head.next.next.next = Node(4)
head.next.next.next.next = Node(5)
 
printlist(head) #print original list
rearrange(head) #rearrange list as per ques
print()
printlist(head) #print modified list
 
# This code is contributed by ab2127

C#

// C# program to rearrange link list in place
using System;
 
// Linked List Class
public class LinkedList {
 
    Node head; // head of the list
 
    /* Node Class */
    class Node {
 
        public int data;
        public Node next;
 
        // Constructor to create a new node
        public Node(int d)
        {
            data = d;
            next = null;
        }
    }
 
    void printlist(Node node)
    {
        if (node == null) {
            return;
        }
        while (node != null) {
            Console.Write(node.data + " -> ");
            node = node.next;
        }
    }
 
    Node reverselist(Node node)
    {
        Node prev = null, curr = node, next;
        while (curr != null) {
            next = curr.next;
            curr.next = prev;
            prev = curr;
            curr = next;
        }
        node = prev;
        return node;
    }
 
    void rearrange(Node node)
    {
 
        // 1) Find the middle point using
        // tortoise and hare method
        Node slow = node, fast = slow.next;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
 
        // 2) Split the linked list in two halves
        // node1, head of first half 1 -> 2 -> 3
        // node2, head of second half 4 -> 5
        Node node1 = node;
        Node node2 = slow.next;
        slow.next = null;
 
        // 3) Reverse the second half, i.e., 5 -> 4
        node2 = reverselist(node2);
 
        // 4) Merge alternate nodes
        node = new Node(0); // Assign dummy Node
 
        // curr is the pointer to this dummy Node, which
        // will be used to form the new list
        Node curr = node;
        while (node1 != null || node2 != null) {
 
            // First add the element from first list
            if (node1 != null) {
                curr.next = node1;
                curr = curr.next;
                node1 = node1.next;
            }
 
            // Then add the element from second list
            if (node2 != null) {
                curr.next = node2;
                curr = curr.next;
                node2 = node2.next;
            }
        }
 
        // Assign the head of the new list to head pointer
        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.printlist(list.head); // print original list
        list.rearrange(
            list.head); // rearrange list as per ques
        Console.WriteLine("");
        list.printlist(list.head); // print modified list
    }
}
 
/* This code is contributed PrinciRaj1992 */

Javascript

<script>
 
// Javascript program to rearrange link list in place
 
// Linked List Class
var head; // head of the list
 
    /* Node Class */
     class Node {
 
// Constructor to create a new node
constructor(d) {
    this.data = d;
    this.next = null;
}
 
    }
 
    function printlist(node) {
        if (node == null) {
            return;
        }
        while (node != null) {
            document.write(node.data + " -> ");
            node = node.next;
        }
    }
 
    function reverselist(node) {
        var prev = null, curr = node, next;
        while (curr != null) {
            next = curr.next;
            curr.next = prev;
            prev = curr;
            curr = next;
        }
        node = prev;
        return node;
    }
 
    function rearrange(node) {
 
        // 1) Find the middle point using tortoise and hare
        // method
        var slow = node, fast = slow.next;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
 
        // 2) Split the linked list in two halves
        // node1, head of first half 1 -> 2 -> 3
        // node2, head of second half 4 -> 5
        var node1 = node;
        var node2 = slow.next;
        slow.next = null;
 
        // 3) Reverse the second half, i.e., 5 -> 4
        node2 = reverselist(node2);
 
        // 4) Merge alternate nodes
        node = new Node(0); // Assign dummy Node
 
        // curr is the pointer to this dummy Node, which
        // will be used to form the new list
        var curr = node;
        while (node1 != null || node2 != null) {
 
            // First add the element from first list
            if (node1 != null) {
                curr.next = node1;
                curr = curr.next;
                node1 = node1.next;
            }
 
            // Then add the element from second list
            if (node2 != null) {
                curr.next = node2;
                curr = curr.next;
                node2 = node2.next;
            }
        }
 
        // Assign the head of the new list to head pointer
        node = node.next;
    }
 
     
        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);
 
        printlist(head); // print original list
        rearrange(head); // rearrange list as per ques
        document.write("<br/>");
        printlist(head); // print modified list
 
// This code contributed by gauravrajput1
</script>

C

// C program to rearrange a linked list in-place
#include <stdio.h>
#include <stdlib.h>
 
// Linkedlist Node structure
typedef struct Node {
    int data;
    struct Node* next;
} Node;
 
// Function to create newNode in a linkedlist
Node* newNode(int key)
{
    Node* temp = (Node*)malloc(sizeof(Node));
    temp->data = key;
    temp->next = NULL;
    return temp;
}
 
// Function to reverse the linked list
void reverselist(Node** head)
{
    // Initialize prev and current pointers
    Node *prev = NULL, *curr = *head, *next;
    while (curr) {
        next = curr->next;
        curr->next = prev;
        prev = curr;
        curr = next;
    }
    *head = prev;
}
 
// Function to print the linked list
void printlist(Node* head)
{
    while (head != NULL) {
        printf("%d  ", head->data);
        if (head->next)
            printf("-> ");
        head = head->next;
    }
    printf("\n");
}
// Function to rearrange a linked list
void rearrange(Node** head)
{
    // 1) Find the middle point using tortoise and hare
    // method
    Node *slow = *head, *fast = slow->next;
    while (fast && fast->next) {
        slow = slow->next;
        fast = fast->next->next;
    }
    // 2) Split the linked list in two halves
    // head1, head of first half    1 -> 2
    // head2, head of second half   3 -> 4
    Node* head1 = *head;
    Node* head2 = slow->next;
    slow->next = NULL;
    // 3) Reverse the second half, i.e.,  4 -> 3
    reverselist(&head2);
    // 4) Merge alternate nodes
    *head = newNode(0); // Assign dummy Node
    // curr is the pointer to this dummy Node, which will
    // be used to form the new list
    Node* curr = *head;
    while (head1 || head2) {
        // First add the element from list
        if (head1) {
            curr->next = head1;
            curr = curr->next;
            head1 = head1->next;
        }
        // Then add the element from the second list
        if (head2) {
            curr->next = head2;
            curr = curr->next;
            head2 = head2->next;
        }
    }
    // Assign the head of the new list to head pointer
    *head = (*head)->next;
}
 
// Driver program
int main()
{
    Node* head = newNode(1);
    head->next = newNode(2);
    head->next->next = newNode(3);
    head->next->next->next = newNode(4);
    head->next->next->next->next = newNode(5);
 
    printlist(head); // Print original list
    rearrange(&head); // Modify the list
    printlist(head); // Print modified list
    return 0;
}
 
// This code is contributed by Aditya Kumar (adityakumar129)

Output

1 -> 2 -> 3 -> 4 -> 5 
1 -> 5 -> 2 -> 4 -> 3

Time Complexity: O(n)
Auxiliary Space: O(1)

Another Approach: (Using recursion)

Hold a pointer to the head node and go till the last node using recursion
Once the last node is reached, start swapping the last node to the next of head node
Move the head pointer to the next node
Repeat this until the head and the last node meet or come adjacent to each other
Once the Stop condition met, we need to discard the left nodes to fix the loop created in the list while swapping nodes.

C++

// C++ implementation
 
#include <bits/stdc++.h>
using namespace std;
 
// Creating the structure for node
struct Node {
    int data;
    struct Node* next;
};
 
// Function to create newNode in a linkedlist
Node* newNode(int key)
{
    Node* temp = new Node;
    temp->data = key;
    temp->next = NULL;
    return temp;
}
 
// Function to print the list
void printlist(Node* head)
{
    while (head) {
        cout << head->data;
        if (head->next)
            cout << "->";
        head = head->next;
    }
    cout << endl;
}
 
// Function to rearrange
void rearrange(Node** head, Node* last)
{
    if (!last)
        return;
    // Recursive call
    rearrange(head, last->next);
    // (*head)->next will be set to NULL after
    // rearrangement. Need not do any operation further Just
    // return here to come out of recursion
    if (!(*head)->next)
        return;
 
    // Rearrange the list until both head and last meet or
    // next to each other.
    if ((*head) != last && (*head)->next != last) {
        Node* tmp = (*head)->next;
        (*head)->next = last;
        last->next = tmp;
        *head = tmp;
    }
    else {
        if ((*head) != last)
            *head = (*head)->next;
        (*head)->next = NULL;
    }
}
 
// Drivers Code
int main()
{
    Node* head = newNode(1);
    head->next = newNode(2);
    head->next->next = newNode(3);
    head->next->next->next = newNode(4);
    head->next->next->next->next = newNode(5);
 
    // Print original list
    printlist(head);
 
    Node* tmp = head;
 
    // Modify the list
    rearrange(&tmp, head);
 
    // Print modified list
    printlist(head);
    return 0;
}
 
// This code is contributed by Aditya Kumar (adityakumar129)

Java

// Java implementation
import java.io.*;
 
// Creating the structure for node
class Node {
    int data;
    Node next;
 
    // Function to create newNode in a linkedlist
    Node(int key)
    {
        data = key;
        next = null;
    }
}
class GFG {
 
    Node left = null;
 
    // Function to print the list
    void printlist(Node head)
    {
        while (head != null) {
            System.out.print(head.data + " ");
            if (head.next != null) {
                System.out.print("->");
            }
            head = head.next;
        }
        System.out.println();
    }
 
    // Function to rearrange
    void rearrange(Node head)
    {
 
        if (head != null) {
            left = head;
            reorderListUtil(left);
        }
    }
 
    void reorderListUtil(Node right)
    {
 
        if (right == null) {
            return;
        }
 
        reorderListUtil(right.next);
 
        // we set left = null, when we reach stop condition,
        // so no processing required after that
        if (left == null) {
            return;
        }
 
        // Stop condition: odd case : left = right, even
        // case : left.next = right
        if (left != right && left.next != right) {
            Node temp = left.next;
            left.next = right;
            right.next = temp;
            left = temp;
        }
        else { // stop condition , set null to left nodes
            if (left.next == right) {
                left.next.next = null; // even case
                left = null;
            }
            else {
                left.next = null; // odd case
                left = null;
            }
        }
    }
 
    // Drivers Code
    public static void main(String[] args)
    {
 
        Node 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);
       
        GFG gfg = new GFG();
 
        // Print original list
        gfg.printlist(head);       
 
        // Modify the list
        gfg.rearrange(head);
 
        // Print modified list
        gfg.printlist(head);
    }
}
 
// This code is contributed by Vishal Singh

Python3

# Python3 implementation
class Node:
     
    def __init__(self, key):
         
        self.data = key
        self.next = None
 
left = None
 
# Function to print the list
def printlist(head):
     
    while (head != None):
        print(head.data, end = " ")
        if (head.next != None):
            print("->", end = "")
             
        head = head.next
         
    print()
 
# Function to rearrange
def rearrange(head):
     
    global left
    if (head != None):
        left = head
        reorderListUtil(left)
 
def reorderListUtil(right):
     
    global left
    if (right == None):
        return
     
    reorderListUtil(right.next)
     
    # We set left = null, when we reach stop
    # condition, so no processing required
    # after that
    if (left == None):
        return
     
    # Stop condition: odd case : left = right, even
    # case : left.next = right
    if (left != right and left.next != right):
        temp = left.next
        left.next = right
        right.next = temp
        left = temp
    else:
         
        # Stop condition , set null to left nodes
        if (left.next == right):
             
            # Even case
            left.next.next = None
            left = None
        else:
             
            # Odd case
            left.next = None
            left = None
 
# Driver code
head = Node(1)
head.next = Node(2)
head.next.next = Node(3)
head.next.next.next = Node(4)
head.next.next.next.next = Node(5)
 
# Print original list
printlist(head)
 
#  Modify the list
rearrange(head)
 
# Print modified list
printlist(head)
 
# This code is contributed by patel2127

C#

// C# implementation
using System;
 
// Creating the structure for node
public class Node
{
    public int data;
    public Node next;
     
    // Function to create newNode
    // in a linkedlist
    public Node(int key)
    {
        data = key;
        next = null;
    }
}
 
class GFG{
     
Node left = null;
 
// Function to print the list
void printlist(Node head)
{
    while (head != null)
    {
        Console.Write(head.data + " ");
        if (head.next != null)
        {
            Console.Write("->");
        }
        head = head.next;
    }
    Console.WriteLine();
}
 
// Function to rearrange
void rearrange(Node head)
{
 
    if (head != null)
    {
        left = head;
        reorderListUtil(left);
    }
}
 
void reorderListUtil(Node right)
{
 
    if (right == null)
    {
        return;
    }
 
    reorderListUtil(right.next);
 
    // We set left = null, when we reach stop
    // condition, so no processing required
    // after that
    if (left == null)
    {
        return;
    }
 
    // Stop condition: odd case : left = right, even
    // case : left.next = right
    if (left != right && left.next != right)
    {
        Node temp = left.next;
        left.next = right;
        right.next = temp;
        left = temp;
    }
    else
    {
         
        // Stop condition , set null to left nodes
        if (left.next == right)
        {
             
            // Even case
            left.next.next = null;
            left = null;
        }
        else
        {
             
            // Odd case
            left.next = null;
            left = null;
        }
    }
}
 
// Driver Code
static public void Main()
{
    Node 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);
    
    GFG gfg = new GFG();
 
    // Print original list
    gfg.printlist(head);       
 
    // Modify the list
    gfg.rearrange(head);
 
    // Print modified list
    gfg.printlist(head);
}
}
 
// This code is contributed by rag2127

Javascript

<script>
// javascript implementation// Creating the structure for node
class Node {
 
    // Function to create newNode in a linkedlist
    constructor(val) {
        this.data = val;
        this.next = null;
    }
}
    var left = null;
 
    // Function to print the list
    function printlist(head) {
        while (head != null) {
            document.write(head.data + " ");
            if (head.next != null) {
                document.write("->");
            }
            head = head.next;
        }
        document.write("<br/>");
    }
 
    // Function to rearrange
    function rearrange(head) {
 
        if (head != null) {
            left = head;
            reorderListUtil(left);
        }
    }
 
    function reorderListUtil(right) {
 
        if (right == null) {
            return;
        }
 
        reorderListUtil(right.next);
 
        // we set left = null, when we reach stop condition,
        // so no processing required after that
        if (left == null) {
            return;
        }
 
        // Stop condition: odd case : left = right, even
        // case : left.next = right
        if (left != right && left.next != right) {
    var temp = left.next;
            left.next = right;
            right.next = temp;
            left = temp;
        } else { // stop condition , set null to left nodes
            if (left.next == right) {
                left.next.next = null; // even case
                left = null;
            } else {
                left.next = null; // odd case
                left = null;
            }
        }
    }
 
    // Drivers 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);
 
 
        // Print original list
        printlist(head);
 
        // Modify the list
        rearrange(head);
 
        // Print modified list
        printlist(head);
 
// This code contributed by aashish1995
</script>

C

// C implementation
#include <stdio.h>
#include <stdlib.h>
 
// Creating the structure for node
typedef struct Node {
    int data;
    struct Node* next;
}Node;
 
// Function to create newNode in a linkedlist
Node* newNode(int key)
{
    Node* temp = malloc(sizeof(Node));
    temp->data = key;
    temp->next = NULL;
    return temp;
}
 
// Function to print the list
void printlist(Node* head)
{
    while (head) {
        printf("%d ", head->data);
        if (head->next)
            printf("->");
        head = head->next;
    }
    printf("\n");
}
 
// Function to rearrange
void rearrange(Node** head, Node* last)
{
 
    if (!last)
        return;
 
    // Recursive call
    rearrange(head, last->next);
 
    // (*head)->next will be set to NULL
    // after rearrangement.
    // Need not do any operation further
    // Just return here to come out of recursion
    if (!(*head)->next)
        return;
 
    // Rearrange the list until both head
    // and last meet or next to each other.
    if ((*head) != last && (*head)->next != last) {
        Node* tmp = (*head)->next;
        (*head)->next = last;
        last->next = tmp;
        *head = tmp;
    }
    else {
        if ((*head) != last)
            *head = (*head)->next;
        (*head)->next = NULL;
    }
}
 
// Drivers Code
int main()
{
    Node* head = newNode(1);
    head->next = newNode(2);
    head->next->next = newNode(3);
    head->next->next->next = newNode(4);
    head->next->next->next->next = newNode(5);
 
    // Print original list
    printlist(head);
 
    Node* tmp = head;
 
    // Modify the list
    rearrange(&tmp, head);
 
    // Print modified list
    printlist(head);
    return 0;
}
 
// This code is contributed by Aditya Kumar (adityakumar129)