# Rearrange a linked list in to alternate first and last element

• Difficulty Level : Hard
• Last Updated : 30 Sep, 2021

Given a linked list. arrange the linked list in manner of alternate first and last element.

Examples:

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```Input : 1->2->3->4->5->6->7->8
Output :1->8->2->7->3->6->4->5

Input :10->11->15->13
Output :10->13->11->15```

We have discussed three different solution in Rearrange a given linked list in-place.
In this post a different Deque based solution is discussed.

Method
1) Create an empty deque
2) Insert all element from the linked list to the deque
3) Insert the element back to the linked list from deque in alternate fashion i.e first then last and so on

## C++

 `// CPP program to rearrange a linked list in given manner``#include ``using` `namespace` `std;` `/* Link list node */``struct` `Node {``    ``int` `data;``    ``struct` `Node* next;``};` `/* Function to reverse the linked list */``void` `arrange(``struct` `Node* head)``{``    ``struct` `Node* temp = head;``    ``deque<``int``> d; ``// defining a deque` `    ``// push all the elements of linked list in to deque``    ``while` `(temp != NULL) {``        ``d.push_back(temp->data);``        ``temp = temp->next;``    ``}` `    ``// Alternatively push the first and last elements``    ``// from deque to back to the linked list and pop``    ``int` `i = 0;``    ``temp = head;``    ``while` `(!d.empty()) {``        ``if` `(i % 2 == 0) {``            ``temp->data = d.front();``            ``d.pop_front();``        ``}``        ``else` `{``            ``temp->data = d.back();``            ``d.pop_back();``        ``}``        ``i++;``        ``temp = temp->next; ``// increase temp``    ``}``}` `/*UTILITY FUNCTIONS*/``/* Push a node to linked list. Note that this function``changes the head */``void` `push(``struct` `Node** head_ref, ``char` `new_data)``{``    ``/* allocate node */``    ``struct` `Node* new_node = (``struct` `Node*)``malloc``(``sizeof``(``struct` `Node));` `    ``/* put in the data */``    ``new_node->data = new_data;` `    ``/* link the old list off the new node */``    ``new_node->next = (*head_ref);` `    ``/* move the head to pochar to the new node */``    ``(*head_ref) = new_node;``}` `// printing the linked list``void` `printList(``struct` `Node* head)``{``    ``struct` `Node* temp = head;``    ``while` `(temp != NULL) {``        ``printf``(``"%d  "``, temp->data);``        ``temp = temp->next;``    ``}``}` `/* Driver program to test above function*/``int` `main()``{``    ``// Let us create linked list 1->2->3->4``    ``struct` `Node* head = NULL;` `    ``push(&head, 5);``    ``push(&head, 4);``    ``push(&head, 3);``    ``push(&head, 2);``    ``push(&head, 1);``    ``cout << ``"Given linked list\t"``;``    ``printList(head);``    ``arrange(head);``    ``cout << ``"\nAfter rearrangement\t"``;``    ``printList(head);``    ``return` `0;``}`

## Java

 `// Java program to rearrange a linked list in given manner``import` `java.util.*;``import` `java.lang.*;``import` `java.io.*;` `class` `GFG``{``    ``/* Link list node */``    ``static` `class` `Node``    ``{``        ``int` `data;``        ``Node next;``        ``Node(``int` `data)``        ``{``            ``this``.data = data;``            ``next = ``null``;``        ``}``    ``}``    ` `    ``// printing the linked list``    ``static` `void` `printList(Node head)``    ``{``        ``Node temp = head;``        ``while` `(temp != ``null``)``        ``{``            ``System.out.print(temp.data + ``" "``);``            ``temp = temp.next;``        ``}``    ``}` `    ``/* Function to reverse the linked list */``    ``static` `void` `arrange(Node head)``    ``{``        ``// defining a deque``        ``Deque deque = ``new` `ArrayDeque<>();``        ``Node temp = head;``        ` `        ``// push all the elements of linked list in to deque``        ``while``(temp != ``null``)``        ``{``            ``deque.addLast(temp.data);``            ``temp = temp.next;``        ``}``        ``temp = head;``        ``int` `i = ``0``;``        ` `        ``// Alternatively push the first and last elements``        ``// from deque to back to the linked list and pop``        ``while``(!deque.isEmpty())``        ``{``            ``if``(i % ``2` `== ``0``)``            ``{``                ``temp.data = deque.removeFirst();``            ``}``            ``else``            ``{``                ``temp.data = deque.removeLast();``            ``}``            ``i++;``            ``temp = temp.next;``        ``}``    ``}` `    ``// Driver code``    ``public` `static` `void` `main (String[] args)``    ``{``        ``// Let us create linked list 1->2->3->4->5``        ``Node head = ``null``;``        ` `        ``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``);` `        ``System.out.println(``"Given linked list"``);``        ``printList(head);``        ``arrange(head);``        ``System.out.println(``"\nAfter rearrangement"``);``        ``printList(head);``    ``}``}` `// This code is contributed by nobody_cares.`

## Python

 `# Python program to rearrange``# a linked list in given manner` `# Link list node``class` `Node:``    ` `    ``def` `__init__(``self``, data):``        ``self``.data ``=` `data``        ``self``.``next` `=` `None` `# Function to reverse the linked list``def` `arrange( head):` `    ``temp ``=` `head``    ` `    ``# defining a deque``    ``d ``=` `[]``    ` `    ``# push all the elements of linked list in to deque``    ``while` `(temp !``=` `None``) :``        ``d.append(temp.data)``        ``temp ``=` `temp.``next``    ` `    ``# Alternatively push the first and last elements``    ``# from deque to back to the linked list and pop``    ``i ``=` `0``    ``temp ``=` `head``    ``while` `(``len``(d) > ``0``) :``        ``if` `(i ``%` `2` `=``=` `0``) :``            ``temp.data ``=` `d[``0``]``            ``d.pop(``0``)``        ` `        ``else` `:``            ``temp.data ``=` `d[``-``1``]``            ``d.pop()``        ` `        ``i ``=` `i ``+` `1` `        ``# increase temp``        ``temp ``=` `temp.``next``        ` `    ``return` `head``    ` `# UTILITY FUNCTIONS``# Push a node to linked list. Note that this function``# changes the head``def` `push( head_ref, new_data):` `    ``# allocate node``    ``new_node ``=` `Node(``0``)` `    ``# put in the data``    ``new_node.data ``=` `new_data` `    ``# link the old list off the new node``    ``new_node.``next` `=` `(head_ref)` `    ``# move the head to pochar to the new node``    ``(head_ref) ``=` `new_node``    ``return` `head_ref` `# printing the linked list``def` `printList( head):` `    ``temp ``=` `head``    ``while` `(temp !``=` `None``) :``        ``print``( temp.data,end``=``" "``)``        ``temp ``=` `temp.``next` `# Driver program to test above function` `# Let us create linked list 1.2.3.4``head ``=` `None` `head ``=` `push(head, ``5``)``head ``=` `push(head, ``4``)``head ``=` `push(head, ``3``)``head ``=` `push(head, ``2``)``head ``=` `push(head, ``1``)``print``(``"Given linked list\t"``)``printList(head)``head ``=` `arrange(head)``print``( ``"\nAfter rearrangement\t"``)``printList(head)` `# This code is contributed by Arnab Kundu`

## Javascript

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Output:

```Given linked list    1 2 3 4 5
After rearrangement    1 5 2 4 3```

Time Complexity : O(n)

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