Javascript Program For Rearranging A Linked List In Zig-Zag Fashion

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

Examples:

Input:  1->2->3->4
Output: 1->3->2->4 
Explanation: 1 and 3 should come first before 2 and 4 in
             zig-zag fashion, So resultant linked-list 
             will be 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 a number of elements in the linked list.

An efficient approach that 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 to this for a detailed explanation of the swapping order. 

Javascript

<script>
// Javascript program to arrange
// linked list in zigzag fashion

// Link list Node
class Node
{
    constructor()
    {
        this.data = 0;
        this.next = null;
    }
}

var head = null;
var temp = 0;

// This function distributes
// the Node in zigzag fashion
function zigZagList(head)
{
    // If flag is true, then
    // next node should be greater
    // in the desired output.
    var flag = true;

    // Traverse linked list starting from
    // head.
    var current = head;
    while (current != null &&
           current.next != null)
    {
        // "<" relation expected
        if (flag == true)
        {
            /* 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;
            }
        }

        // ">" relation expected
        else
        {
            /* 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
Function push(new_data)
{
    // Allocate Node
    var new_Node = new Node();

    // Put in the data
    new_Node.data = new_data;

    // Link the old list of the
    // new Node
    new_Node.next = (head);

    // Move the head to point to
    // the new Node
    (head) = new_Node;
}

// Function to print linked list
function printList(node)
{
    while (node != null)
    {
        document.write(node.data + "->");
        node = node.next;
    }
    document.write("NULL<br/>");
}

// Driver code
// 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);

document.write(
         "Given linked list <br/>");
printList(head);
zigZagList(head);
document.write(
         "Zig Zag Linked list <br/>");
printList(head);
// This code is contributed by gauravrajput1
</script>

Output:

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

Time Complexity: O(N), as we are using a loop for traversing the linked list.

Auxiliary Space: O(1), as we are not using extra space.

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

This can be also be done recursively. The idea remains the same, let us suppose the value of the flag determines the condition we need to check for comparing the current element. So, if the flag is 0 (or false) the current element should be smaller than the next and if the flag is 1 ( or true ) then the current element should be greater than the next. If not, swap the values of nodes.

Javascript

<script>
// Javascript program for the
// above approach
// Node class
class Node
{
    constructor(val)
    {
        this.data = val;
        this.next = null;
    }
}

var head;

// Print Linked List
function printLL()
{
    var t = head;
    while (t != null)
    {
        document.write(t.data +
                       " ->");
        t = t.next;
    }
    document.write();
}

// Swap both nodes
function swap(a, b)
{
    if (a == null ||
        b == null)
        return;
    var temp = a.data;
    a.data = b.data;
    b.data = temp;
}

// Rearrange the linked list
// in zig zag way
function zigZag(node, flag)
{
    if (node == null ||
        node.next == null)
    {
        return node;
    }
    if (flag == 0)
    {
        if (node.data >
            node.next.data)
        {
            swap(node, node.next);
        }
        return zigZag(node.next, 1);
    }
    else
    {
        if (node.data <
            node.next.data)
        {
            swap(node, node.next);
        }
        return zigZag(node.next, 0);
    }
}

// Driver Code
head = new Node(11);
head.next = new Node(15);
head.next.next = new Node(20);
head.next.next.next = new Node(5);
head.next.next.next.next =
new Node(10);
printLL();

// 0 means the current element
// should be smaller than the next
var flag = 0;
zigZag(head, flag);
document.write(
"<br/>LL in zig zag fashion : <br/>");
printLL();
// This code is contributed by umadevi9616
</script>

Output:

11 ->15 ->20 ->5 ->10 ->
LL in zig zag fashion : 
11 ->20 ->5 ->15 ->10 ->

Complexity Analysis:

• Time Complexity: O(n). 
  Traversal of the list is done only once, and it has ‘n’ elements.
• Auxiliary Space: O(n). 
  O(n) extra space  due to recursive calls

Please refer complete article on Rearrange a Linked List in Zig-Zag fashion for more details!