Rearrange a Linked List in Zig-Zag fashion

https://www.geeksforgeeks.org/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 linked list. Examples :

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

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 (< or >) 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.

// 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 */

    }

}

https://www.tutorialcup.com/linked-list/rearrange-linked-list-zig-zag.htm

https://www.geeksforgeeks.org/rearrange-a-linked-list-in-zig-zag-fashion-set-2/

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.

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;

}

X. https://www.geeksforgeeks.org/convert-array-into-zig-zag-fashion/

Given an array of DISTINCT elements, rearrange the elements of array in zig-zag fashion in O(n) time. The converted array should be in form a < b > c < d > e < f.

Example: 
Input:  arr[] = {4, 3, 7, 8, 6, 2, 1}
Output: arr[] = {3, 7, 4, 8, 2, 6, 1}

We can convert in O(n) time using an Efficient Approach. The idea is to use modified one pass of bubble sort. Maintain a flag for representing which order(i.e. < or >) currently we need. If the current two elements are not in that order then swap those elements otherwise not.
Let us see the main logic using three consecutive elements A, B, C. Suppose we are processing B and C currently and the current relation is ‘<'. But we have B > C. Since current relation is ‘<' previous relation must be '>‘ i.e., A must be greater than B. So, the relation is A > B and B > C. We can deduce A > C. So if we swap B and C then the relation is A > C and C < B. Finally we get the desired order A C B

    static void zigZag()

    {

        // Flag true indicates relation "<" is expected,

        // else ">" is expected.  The first expected relation

        // is "<"

        boolean flag = true;

          

        int temp =0;

       

        for (int i=0; i<=arr.length-2; i++)

        {

            if (flag)  /* "<" relation expected */

            {

                /* If we have a situation like A > B > C,

                   we get A > B < C by swapping B and C */

                if (arr[i] > arr[i+1])

                {

                    // swap

                    temp  = arr[i];

                    arr[i] = arr[i+1];

                    arr[i+1] = temp;

                }

                  

            }

            else /* ">" relation expected */

            {

                /* If we have a situation like A < B < C,

                   we get A < C > B by swapping B and C */

                if (arr[i] < arr[i+1])

                {

                    // swap

                    temp = arr[i];

                    arr[i] = arr[i+1];

                    arr[i+1] = temp;

                }

            }

            flag = !flag; /* flip flag */

        }

    }