Rearrange a given linked list in-place.

Given a singly linked list L0 -> L1 -> … -> Ln-1 -> Ln. Rearrange the nodes in the list so that the new formed list is : L0 -> Ln -> L1 -> Ln-1 -> L2 -> Ln-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 to next of current

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

Better Solution
1) Copy contents of given linked list to a vector.
2) Rearrange given vector by swapping nodes from both ends.
3) Copy the modified vector back to the linked list.
Implementation of this approach : https://ide.geeksforgeeks.org/1eGSEy
Thanks to Arushi Dhamija for suggesting this approach.

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. 

Time Complexity of this solution is O(n).

Below is the implementation of this method.

C++

// C++ program to rearrange a linked list in-place
#include
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;
}

Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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 */

chevron_right


Output:

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

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

Thanks to Gaurav Ahirwar for suggesting the above approach.

Another approach :
1. Take two pointers prev and curr, which hold the addresses of head and head-> next.
2. Compare their data and swap.
After that, a new linked list is formed.
Below is the implementation :

filter_none

edit
close

play_arrow

link
brightness_4
code

// CPP code to rearrange linked list in place
#include <bits/stdc++.h>
#include <iostream>
using namespace std;
  
struct node {
    int data;
    struct node* next;
};
typedef struct node Node;
  
// function for rearranging a linked list with high and low value.
void rearrange(Node* head)
{
    if (head == NULL) // Base case.
        return;
  
    // two pointer variable.
    Node *prev = head, *curr = head->next;
  
    while (curr) {
        // swap function for swapping data.
        if (prev->data > curr->data)
            swap(prev->data, curr->data);
  
        // swap function for swapping data.
        if (curr->next && curr->next->data > curr->data)
            swap(curr->next->data, curr->data);
  
        prev = curr->next;
  
        if (!curr->next)
            break;
        curr = curr->next->next;
    }
}
  
// function to insert a node in the linked list at the beginning.
void push(Node** head, int k)
{
    Node* tem = (Node*)malloc(sizeof(Node));
    tem->data = k;
    tem->next = *head;
    *head = tem;
}
  
// function to display node of linked list.
void display(Node* head)
{
    Node* curr = head;
    while (curr != NULL) {
        printf("%d ", curr->data);
        curr = curr->next;
    }
}
  
// driver code
int main()
{
  
    Node* head = NULL;
  
    // let create a linked list.
    // 9 -> 6 -> 8 -> 3 -> 7
    push(&head, 7);
    push(&head, 3);
    push(&head, 8);
    push(&head, 6);
    push(&head, 9);
  
    rearrange(head);
  
    display(head);
  
    return 0;
}

chevron_right


Output:

6 9 3 8 7

Time Complexity : O(n)
Auxiliary Space : O(1)
Thanks to Aditya for suggesting this approach.

Another Approach: ( Using recursion )

  1. Hold a pointer to the head node and go till the last node using recursion
  2. Once last node is reached, start swapping the last node to the next of head node
  3. Move the head pointer to the next node
  4. Repeat this until the head and last node meet or come adjacent to each other
filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation
#include <stdio.h>
#include <stdlib.h>
  
// Creating the structure for node
struct Node {
    int data;
    struct Node* next;
};
  
// Function to create newNode in a linkedlist
struct Node* newNode(int key)
{
    struct Node* temp = malloc(sizeof(struct Node));
    temp->data = key;
    temp->next = NULL;
    return temp;
}
  
// Function to print the list
void printlist(struct Node* head)
{
    while (head) {
        printf("%d ", head->data);
        if (head->next)
            printf("->");
        head = head->next;
    }
    printf("\n");
}
  
// Function to rearrange
void rearrange(struct Node** head, struct 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) {
        struct 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()
{
    struct 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);
  
    struct Node* tmp = head;
  
    // Modify the list
    rearrange(&tmp, head);
  
    // Print modified list
    printlist(head);
    return 0;
}

chevron_right


Output:

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

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above



My Personal Notes arrow_drop_up



Article Tags :
Practice Tags :


10


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.