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 linked list. Note that it is not allowed to swap data.

Examples:

Input: 1->2->3->4 Output: 1->3->2->4 Input: 11->15->20->5->10 Output: 11->20->5->15->10

**Approach: **

A solution that converts given list into zigzag form is discussed in previous post. The solution discussed performs conversion by swapping data of nodes. Swapping data of nodes may be expensive in many situations when the data contains many fields. In this post, a solution that performs conversion by swapping links is discussed.

The idea is to traverse the given linked list and check if current node maintains the zigzag order or not. To check if given node maintains zigzag order or not, a variable *ind* is used. If **ind = 0**, then the current node’s data should be less than its adjacent node’s data and if** ind = 1, ** then current node’s data should be greater than its adjacent node’s data. If the current node violates the zigzag order, then swap the position of both nodes. For doing this step, maintain two pointers **prev and next**. *prev* stores previous node of current node and *next* stores new next node of current node. To swap both nodes, the following steps are performed:

- Make next node of current node, the next node of previous node.
- Make the current node next node of its adjacent node.
- Make current node next = next node.

Below is the implementation of above approach:

`// CPP program to arrange linked list in ` `// zigzag fashion ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `/* Link list Node */` `struct` `Node { ` ` ` `int` `data; ` ` ` `struct` `Node* next; ` `}; ` ` ` `// This function converts the Linked list in ` `// zigzag fashion ` `Node* zigZagList(Node* head) ` `{ ` ` ` `if` `(head == NULL || head->next == NULL) { ` ` ` `return` `head; ` ` ` `} ` ` ` ` ` `// to store new head ` ` ` `Node* res = NULL; ` ` ` ` ` `// to traverse linked list ` ` ` `Node* curr = head; ` ` ` ` ` `// to store previous node of current node ` ` ` `Node* prev = NULL; ` ` ` ` ` `// to store new next node of current node ` ` ` `Node* next; ` ` ` ` ` `// to check if current element should ` ` ` `// be less than or greater than. ` ` ` `// ind = 0 --> less than ` ` ` `// ind = 1 --> greater than ` ` ` `int` `ind = 0; ` ` ` ` ` `while` `(curr->next) { ` ` ` ` ` `// If elements are not in zigzag fashion ` ` ` `// swap them. ` ` ` `if` `((ind == 0 && curr->data > curr->next->data) ` ` ` `|| (ind == 1 && curr->data < curr->next->data)) { ` ` ` ` ` `if` `(res == NULL) ` ` ` `res = curr->next; ` ` ` ` ` `// Store new next element of current ` ` ` `// node ` ` ` `next = curr->next->next; ` ` ` ` ` `// Previous node of current node will ` ` ` `// now point to next node of current node ` ` ` `if` `(prev) ` ` ` `prev->next = curr->next; ` ` ` ` ` `// Change next pointers of both ` ` ` `// adjacent nodes ` ` ` `curr->next->next = curr; ` ` ` `curr->next = next; ` ` ` ` ` `// Change previous pointer. ` ` ` `if` `(prev) ` ` ` `prev = prev->next; ` ` ` `else` ` ` `prev = res; ` ` ` `} ` ` ` ` ` `// If already in zig zag form, then move ` ` ` `// to next element. ` ` ` `else` `{ ` ` ` `if` `(res == NULL) { ` ` ` `res = curr; ` ` ` `} ` ` ` ` ` `prev = curr; ` ` ` `curr = curr->next; ` ` ` `} ` ` ` ` ` `// Update info whether next element should ` ` ` `// be less than or greater than. ` ` ` `ind = 1 - ind; ` ` ` `} ` ` ` ` ` `return` `res; ` `} ` ` ` `/* 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; ` ` ` `} ` `} ` ` ` `/* 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); ` ` ` ` ` `head = zigZagList(head); ` ` ` ` ` `printf` `(` `"\nZig Zag Linked list \n"` `); ` ` ` `printList(head); ` ` ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

**Output:**

Given linked list 4->3->7->8->6->2->1-> Zig Zag Linked list 3->7->4->8->2->6->1->

**Time Complexity:** O(N)

**Auxiliary Space:** O(1)

## Recommended Posts:

- Rearrange a Linked List in Zig-Zag fashion
- Rearrange a given linked list in-place.
- Rearrange a linked list such that all even and odd positioned nodes are together
- Rearrange a linked list in to alternate first and last element
- Convert singly linked list into circular linked list
- XOR Linked List – A Memory Efficient Doubly Linked List | Set 2
- Merge a linked list into another linked list at alternate positions
- XOR Linked List - A Memory Efficient Doubly Linked List | Set 1
- Modify and Rearrange List
- Check if a linked list is Circular Linked List
- Rearrange a given list such that it consists of alternating minimum maximum elements
- Construct a Maximum Sum Linked List out of two Sorted Linked Lists having some Common nodes
- Partitioning a linked list around a given value and If we don't care about making the elements of the list "stable"
- Length of longest palindrome list in a linked list using O(1) extra space
- Rotate the sub-list of a linked list from position M to N to the right by K places

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