# Rearrange a Linked List in Zig-Zag fashion

Given a linked list, rearrange it such that converted list should be of the form a < b > c < d > e < f .. where a, b, c.. are consecutive data node of the linked list.

Examples:

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

Input:  11->15->20->5->10
Output: 11->20->5->15->10
```

## We strongly recommend that you click here and practice it, before moving on to the solution.

A simple approach to do this, is to sort the linked list using merge sort and then swap alternate, but that requires O(n Log n) time complexity. Here n is number of elements in linked list.

An efficient approach which requires O(n) time is, using a single scan similar to bubble sort and then maintain a flag for representing which order () currently we are. If the current two elements are not in that order then swap those elements otherwise not. Please refer this for detailed explanation of swapping order.

## C++

 `// C++ program to arrange linked ` `// list in zigzag fashion ` `#include ` `using` `namespace` `std; ` ` `  `/* Link list Node */` `struct` `Node { ` `    ``int` `data; ` `    ``struct` `Node* next; ` `}; ` ` `  `// This function distributes the ` `// Node in zigzag fashion ` `void` `zigZagList(Node* head) ` `{ ` `    ``// If flag is true, then next ` `    ``// node should be greater ` `    ``// in the desired output. ` `    ``bool` `flag = ``true``; ` ` `  `    ``// Traverse linked list starting from head. ` `    ``Node* current = head; ` `    ``while` `(current->next != NULL) { ` `        ``if` `(flag) ``/* "<" relation expected */` `        ``{ ` `            ``/* If we have a situation like A > B > C ` `               ``where A, B and C are consecutive Nodes ` `               ``in list we get A > B < C by swapping B ` `               ``and C */` `            ``if` `(current->data > current->next->data) ` `                ``swap(current->data, current->next->data); ` `        ``} ` `        ``else` `/* ">" relation expected */` `        ``{ ` `            ``/* If we have a situation like A < B < C where ` `               ``A, B and C  are consecutive Nodes in list we ` `               ``get A < C > B by swapping B and C */` `            ``if` `(current->data < current->next->data) ` `                ``swap(current->data, current->next->data); ` `        ``} ` ` `  `        ``current = current->next; ` `        ``flag = !flag; ``/* flip flag for reverse checking */` `    ``} ` `} ` ` `  `/* UTILITY FUNCTIONS */` `/* Function to push a Node */` `void` `push(Node** head_ref, ``int` `new_data) ` `{ ` `    ``/* allocate Node */` `    ``struct` `Node* new_Node = ``new` `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 point to the new Node */` `    ``(*head_ref) = new_Node; ` `} ` ` `  `/* Function to print linked list */` `void` `printList(``struct` `Node* Node) ` `{ ` `    ``while` `(Node != NULL) { ` `        ``printf``(``"%d->"``, Node->data); ` `        ``Node = Node->next; ` `    ``} ` `    ``printf``(``"NULL"``); ` `} ` ` `  `/* Driver program to test above function*/` `int` `main(``void``) ` `{ ` `    ``/* Start with the empty list */` `    ``struct` `Node* head = NULL; ` ` `  `    ``// create a list 4 -> 3 -> 7 -> 8 -> 6 -> 2 -> 1 ` `    ``// answer should be -> 3  7  4  8  2  6  1 ` `    ``push(&head, 1); ` `    ``push(&head, 2); ` `    ``push(&head, 6); ` `    ``push(&head, 8); ` `    ``push(&head, 7); ` `    ``push(&head, 3); ` `    ``push(&head, 4); ` ` `  `    ``printf``(``"Given linked list \n"``); ` `    ``printList(head); ` ` `  `    ``zigZagList(head); ` ` `  `    ``printf``(``"\nZig Zag Linked list \n"``); ` `    ``printList(head); ` ` `  `    ``return` `(0); ` `} `

## Java

 `// Java program to arrange ` `// linked list in zigzag fashion ` `class` `GfG { ` ` `  `    ``/* Link list Node */` `    ``static` `class` `Node { ` `        ``int` `data; ` `        ``Node next; ` `    ``} ` `    ``static` `Node head = ``null``; ` `    ``static` `int` `temp = ``0``; ` ` `  `    ``// This function distributes ` `    ``// the Node in zigzag fashion ` `    ``static` `void` `zigZagList(Node head) ` `    ``{ ` `        ``// If flag is true, then ` `        ``// next node should be greater ` `        ``// in the desired output. ` `        ``boolean` `flag = ``true``; ` ` `  `        ``// Traverse linked list starting from head. ` `        ``Node current = head; ` `        ``while` `(current != ``null` `&& current.next != ``null``) { ` `            ``if` `(flag == ``true``) ``/* "<" relation expected */` `            ``{ ` `                ``/* If we have a situation like A > B > C  ` `            ``where A, B and C are consecutive Nodes  ` `            ``in list we get A > B < C by swapping B  ` `            ``and C */` `                ``if` `(current.data > current.next.data) { ` `                    ``temp = current.data; ` `                    ``current.data = current.next.data; ` `                    ``current.next.data = temp; ` `                ``} ` `            ``} ` `            ``else` `/* ">" relation expected */` `            ``{ ` `                ``/* If we have a situation like A < B < C where  ` `            ``A, B and C are consecutive Nodes in list we  ` `            ``get A < C > B by swapping B and C */` `                ``if` `(current.data < current.next.data) { ` `                    ``temp = current.data; ` `                    ``current.data = current.next.data; ` `                    ``current.next.data = temp; ` `                ``} ` `            ``} ` ` `  `            ``current = current.next; ` ` `  `            ``/* flip flag for reverse checking */` `            ``flag = !(flag); ` `        ``} ` `    ``} ` ` `  `    ``/* UTILITY FUNCTIONS */` `    ``/* Function to push a Node */` `    ``static` `void` `push(``int` `new_data) ` `    ``{ ` `        ``/* allocate Node */` `        ``Node new_Node = ``new` `Node(); ` ` `  `        ``/* put in the data */` `        ``new_Node.data = new_data; ` ` `  `        ``/* link the old list off the new Node */` `        ``new_Node.next = (head); ` ` `  `        ``/* move the head to point to the new Node */` `        ``(head) = new_Node; ` `    ``} ` ` `  `    ``/* Function to print linked list */` `    ``static` `void` `printList(Node Node) ` `    ``{ ` `        ``while` `(Node != ``null``) { ` `            ``System.out.print(Node.data + ``"->"``); ` `            ``Node = Node.next; ` `        ``} ` `        ``System.out.println(``"NULL"``); ` `    ``} ` ` `  `    ``/* Driver code*/` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``/* Start with the empty list */` `        ``// Node head = null; ` ` `  `        ``// create a list 4 -> 3 -> 7 -> 8 -> 6 -> 2 -> 1 ` `        ``// answer should be -> 3 7 4 8 2 6 1 ` `        ``push(``1``); ` `        ``push(``2``); ` `        ``push(``6``); ` `        ``push(``8``); ` `        ``push(``7``); ` `        ``push(``3``); ` `        ``push(``4``); ` ` `  `        ``System.out.println(``"Given linked list "``); ` `        ``printList(head); ` ` `  `        ``zigZagList(head); ` ` `  `        ``System.out.println(``"Zig Zag Linked list "``); ` `        ``printList(head); ` `    ``} ` `} ` ` `  `// This code is contributed by ` `// Prerna Saini. `

## Python

 `# Python code to rearrange linked list in zig zac fashion ` ` `  `# Node class  ` `class` `Node:  ` ` `  `    ``# Constructor to initialize the node object  ` `    ``def` `__init__(``self``, data):  ` `        ``self``.data ``=` `data  ` `        ``self``.``next` `=` `None` ` `  ` `  `# This function distributes the Node in zigzag fashion ` `def` `zigZagList(head): ` ` `  `    ``# If flag is true, then next node should be greater ` `    ``# in the desired output. ` `    ``flag ``=` `True` ` `  `    ``# Traverse linked list starting from head. ` `    ``current ``=` `head ` `    ``while` `(current.``next` `!``=` `None``): ` `     `  `        ``if` `(flag): ``# "<" relation expected ` `         `  `            ``# If we have a situation like A > B > C ` `            ``# where A, B and C are consecutive Nodes ` `            ``# in list we get A > B < C by swapping B ` `            ``# and C  ` `            ``if` `(current.data > current.``next``.data): ` `                ``t ``=` `current.data ` `                ``current.data ``=` `current.``next``.data ` `                ``current.``next``.data ``=` `t ` `             `  `         `  `        ``else` `:``# ">" relation expected ` `         `  `            ``# If we have a situation like A < B < C where ` `            ``# A, B and C are consecutive Nodes in list we ` `            ``# get A < C > B by swapping B and C  ` `            ``if` `(current.data < current.``next``.data): ` `                ``t ``=` `current.data ` `                ``current.data ``=` `current.``next``.data ` `                ``current.``next``.data ``=` `t ` `             `  `        ``current ``=` `current.``next` `        ``if``(flag): ` `            ``flag ``=` `False` `# flip flag for reverse checking  ` `        ``else``: ` `            ``flag ``=` `True` `    ``return` `head ` ` `  `# function to insert a Node in  ` `# the linked list at the beginning. ` `def` `push(head, k): ` ` `  `    ``tem ``=` `Node(``0``) ` `    ``tem.data ``=` `k ` `    ``tem.``next` `=` `head ` `    ``head ``=` `tem ` `    ``return` `head ` ` `  `# function to display Node of linked list. ` `def` `display( head): ` ` `  `    ``curr ``=` `head ` `    ``while` `(curr !``=` `None``):  ` `        ``print``( curr.data, ``"->"``, end ``=``" "``) ` `        ``curr ``=` `curr.``next` `     `  `    ``print``(``"None"``) ` ` `  `# Driver code ` ` `  `head ``=` `None` ` `  `# create a list 4 -> 3 -> 7 -> 8 -> 6 -> 2 -> 1 ` `# answer should be -> 3 7 4 8 2 6 1 ` `head ``=` `push(head, ``1``) ` `head ``=` `push(head, ``2``) ` `head ``=` `push(head, ``6``) ` `head ``=` `push(head, ``8``) ` `head ``=` `push(head, ``7``) ` `head ``=` `push(head, ``3``) ` `head ``=` `push(head, ``4``) ` ` `  `print``(``"Given linked list \n"``) ` `display(head) ` ` `  `head ``=` `zigZagList(head) ` ` `  `print``(``"\nZig Zag Linked list \n"``) ` `display(head) ` ` `  `# This code is contributed by Arnab Kundu `

## C#

 `// C# program to arrange ` `// linked list in zigzag fashion ` `using` `System; ` ` `  `class` `GfG { ` ` `  `    ``/* Link list Node */` `    ``class` `Node { ` `        ``public` `int` `data; ` `        ``public` `Node next; ` `    ``} ` `    ``static` `Node head = ``null``; ` `    ``static` `int` `temp = 0; ` ` `  `    ``// This function distributes ` `    ``// the Node in zigzag fashion ` `    ``static` `void` `zigZagList(Node head) ` `    ``{ ` `        ``// If flag is true, then ` `        ``// next node should be greater ` `        ``// in the desired output. ` `        ``bool` `flag = ``true``; ` ` `  `        ``// Traverse linked list starting from head. ` `        ``Node current = head; ` `        ``while` `(current != ``null` `&& current.next != ``null``) { ` `            ``if` `(flag == ``true``) ``/* "<" relation expected */` `            ``{ ` `                ``/* If we have a situation like A > B > C  ` `                ``where A, B and C are consecutive Nodes  ` `                ``in list we get A > B < C by swapping B  ` `                ``and C */` `                ``if` `(current != ``null` `&& current.next != ``null` `&& current.data > current.next.data) { ` `                    ``temp = current.data; ` `                    ``current.data = current.next.data; ` `                    ``current.next.data = temp; ` `                ``} ` `            ``} ` `            ``else` `/* ">" relation expected */` `            ``{ ` `                ``/* If we have a situation like A < B < C where  ` `                ``A, B and C are consecutive Nodes in list we  ` `                ``get A < C > B by swapping B and C */` `                ``if` `(current != ``null` `&& current.next != ``null` `&& current.data < current.next.data) { ` `                    ``temp = current.data; ` `                    ``current.data = current.next.data; ` `                    ``current.next.data = temp; ` `                ``} ` `            ``} ` ` `  `            ``current = current.next; ` ` `  `            ``/* flip flag for reverse checking */` `            ``flag = !(flag); ` `        ``} ` `    ``} ` ` `  `    ``/* UTILITY FUNCTIONS */` `    ``/* Function to push a Node */` `    ``static` `void` `push(``int` `new_data) ` `    ``{ ` `        ``/* allocate Node */` `        ``Node new_Node = ``new` `Node(); ` ` `  `        ``/* put in the data */` `        ``new_Node.data = new_data; ` ` `  `        ``/* link the old list off the new Node */` `        ``new_Node.next = (head); ` ` `  `        ``/* move the head to point to the new Node */` `        ``(head) = new_Node; ` `    ``} ` ` `  `    ``/* Function to print linked list */` `    ``static` `void` `printList(Node Node) ` `    ``{ ` `        ``while` `(Node != ``null``) { ` `            ``Console.Write(Node.data + ``"->"``); ` `            ``Node = Node.next; ` `        ``} ` `        ``Console.WriteLine(``"NULL"``); ` `    ``} ` ` `  `    ``/* Driver code*/` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``/* Start with the empty list */` `        ``// Node head = null; ` ` `  `        ``// create a list 4 -> 3 -> 7 -> 8 -> 6 -> 2 -> 1 ` `        ``// answer should be -> 3 7 4 8 2 6 1 ` `        ``push(1); ` `        ``push(2); ` `        ``push(6); ` `        ``push(8); ` `        ``push(7); ` `        ``push(3); ` `        ``push(4); ` ` `  `        ``Console.WriteLine(``"Given linked list "``); ` `        ``printList(head); ` ` `  `        ``zigZagList(head); ` ` `  `        ``Console.WriteLine(``"Zig Zag Linked list "``); ` `        ``printList(head); ` `    ``} ` `} ` `/* This code is contributed PrinciRaj1992 */`

Output:

```Given linked list
4->3->7->8->6->2->1->NULL

3->7->4->8->2->6->1->NULL
```

In above code, push function pushes the node at the front of the linked list, the code can be easily modified for pushing node at the end of list. Other thing to note is, swapping of data between two nodes is done by swap by value not swap by links for simplicity, for swap by links technique please see this.

Complexity Analysis:

• Time Complexity: O(n).
Traversal of list is done only once and it has ‘n’ elements.
• Auxiliary Space: O(1).
No use of extra data structure for storing values.

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