Javascript Program For Rearranging A Linked List Such That All Even And Odd Positioned Nodes Are Together

Last Updated : 04 Sep, 2024

Rearrange a linked list in such a way that all odd position nodes are together and all even positions node are together,

Examples:

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

Input:   10->22->30->43->56->70
Output:  10->30->56->22->43->70

The important thing in this question is to make sure that all below cases are handled

Empty linked list.
A linked list with only one node.
A linked list with only two nodes.
A linked list with an odd number of nodes.
A linked list with an even number of nodes.

The below program maintains two pointers 'odd' and 'even' for current nodes at odd and even positions respectively. We also store the first node of the even linked list so that we can attach the even list at the end of the odd list after all odd and even nodes are connected together in two different lists.

JavaScript

// Javascript program to rearrange a linked list 
// in such a way that all odd positioned  
// node are stored before all even 
// positioned nodes 

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

// A utility function to create 
// a new node
function newNode(key) {
    let temp = new Node();
    temp.data = key;
    temp.next = null;
    return temp;
}

// Rearranges given linked list
// such that all even positioned
// nodes are before odd positioned.
// Returns new head of linked List.
function rearrangeEvenOdd(head) {
    // Corner case
    if (head == null)
        return null;

    // Initialize first nodes of even 
    // and odd lists
    let odd = head;
    let even = head.next;

    // Remember the first node of even 
    // list so that we can connect the 
    // even list at the end of odd list.
    let evenFirst = even;

    while (1 == 1) {
        // If there are no more nodes,
        // then connect first node of even
        // list to the last node of odd list
        if (odd == null || even == null ||
            (even.next) == null) {
            odd.next = evenFirst;
            break;
        }

        // Connecting odd nodes
        odd.next = even.next;
        odd = even.next;

        // If there are NO more even nodes
        // after current odd.
        if (odd.next == null) {
            even.next = null;
            odd.next = evenFirst;
            break;
        }

        // Connecting even nodes
        even.next = odd.next;
        even = odd.next;
    }
    return head;
}

// A utility function to point 
// a linked list
function printlist(node) {
    while (node != null) {
        console.log(node.data);
        node = node.next;
    }
}

// Driver code
let head = newNode(1);
head.next = newNode(2);
head.next.next = newNode(3);
head.next.next.next = newNode(4);
head.next.next.next.next = newNode(5);

console.log("Given Linked List");
printlist(head);

head = rearrangeEvenOdd(head);

console.log("Modified Linked List");
printlist(head);
// This code is contributed by gauravrajput1

Output:

Given Linked List
1->2->3->4->5->NULL
Modified Linked List
1->3->5->2->4->NULL

Complexity Analysis:

Time Complexity: O(n) where n is the number of nodes in the given linked list.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

Please refer complete article on Rearrange a linked list such that all even and odd positioned nodes are together for more details!