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; }

**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)

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