# 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 ```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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

 `// 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 `

## 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 */`

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 :
2. Compare their data and swap.
After that, a new linked list is formed.
Below is the implementation :

## C++

 `// CPP code to rearrange linked list in place ` `#include ` `#include ` `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; ` `} `

## Java

 `// Java code to rearrange linked list in place ` `class` `Geeks ` `{ ` `     `  `static` `class` `Node  ` `{ ` `    ``int` `data; ` `    ``Node next; ` `} ` ` `  ` `  `// function for rearranging a linked list ` `// with high and low value. ` `static` `Node rearrange(Node head) ` `{ ` `    ``if` `(head == ``null``) ``// Base case. ` `        ``return` `null``; ` ` `  `    ``// two pointer variable. ` `    ``Node prev = head, curr = head.next; ` ` `  `    ``while` `(curr != ``null``)  ` `    ``{ ` `        ``// swap function for swapping data. ` `        ``if` `(prev.data > curr.data) ` `        ``{ ` `            ``int` `t = prev.data; ` `            ``prev.data = curr.data; ` `            ``curr.data = t; ` `        ``} ` ` `  `        ``// swap function for swapping data. ` `        ``if` `(curr.next != ``null` `&& curr.next.data > curr.data) ` `        ``{ ` `            ``int` `t = curr.next.data; ` `            ``curr.next.data = curr.data; ` `            ``curr.data = t; ` `        ``} ` ` `  `        ``prev = curr.next; ` ` `  `        ``if` `(curr.next == ``null``) ` `            ``break``; ` `        ``curr = curr.next.next; ` `    ``} ` `    ``return` `head; ` `} ` ` `  `// function to insert a Node in  ` `// the linked list at the beginning. ` `static` `Node push(Node head, ``int` `k) ` `{ ` `    ``Node tem = ``new` `Node(); ` `    ``tem.data = k; ` `    ``tem.next = head; ` `    ``head = tem; ` `    ``return` `head; ` `} ` ` `  `// function to display Node of linked list. ` `static` `void` `display(Node head) ` `{ ` `    ``Node curr = head; ` `    ``while` `(curr != ``null``) ` `    ``{ ` `        ``System.out.printf(``"%d "``, curr.data); ` `        ``curr = curr.next; ` `    ``} ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String args[]) ` `{ ` ` `  `    ``Node head = ``null``; ` ` `  `    ``// let create a linked list. ` `    ``// 9 . 6 . 8 . 3 . 7 ` `    ``head = push(head, ``7``); ` `    ``head = push(head, ``3``); ` `    ``head = push(head, ``8``); ` `    ``head = push(head, ``6``); ` `    ``head = push(head, ``9``); ` ` `  `    ``head = rearrange(head); ` ` `  `    ``display(head); ` ` `  `} ` `} ` ` `  `// This code is contributed by Arnab Kundu `

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

 `// C++ implementation ` `#include ` `#include ` ` `  `// 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; ` `} `

Output:

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