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Javascript Program For Alternating Split Of A Given Singly Linked List- Set 1

Last Updated : 23 Jul, 2025
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Write a function AlternatingSplit() that takes one list and divides up its nodes to make two smaller lists 'a' and 'b'. The sublists should be made from alternating elements in the original list. So if the original list is 0->1->0->1->0->1 then one sublist should be 0->0->0 and the other should be 1->1->1.

Method (Using Dummy Nodes): 

Here is an approach that builds the sub-lists in the same order as the source list. The code uses temporary dummy header nodes for the 'a' and 'b' lists as they are being built. Each sublist has a "tail" pointer that points to its current last node — that way new nodes can be appended to the end of each list easily. The dummy nodes give the tail pointers something to point to initially. The dummy nodes are efficient in this case because they are temporary and allocated in the stack. Alternately, local "reference pointers" (which always point to the last pointer in the list instead of to the last node) could be used to avoid Dummy nodes.

JavaScript 
// Node class to create a new node in the linked list
class Node {
    constructor(data) {
        this.data = data;
        this.next = null;
    }
}

// Linked List class
class LinkedList {
    constructor() {
        this.head = null;
    }

    // Method to add a new node at the end of the list
    append(data) {
        let newNode = new Node(data);
        if (this.head === null) {
            this.head = newNode;
        } else {
            let current = this.head;
            while (current.next !== null) {
                current = current.next;
            }
            current.next = newNode;
        }
    }

    // Method to print the linked list
    printList() {
        let current = this.head;
        let result = '';
        while (current !== null) {
            result += current.data + ' ';
            current = current.next;
        }
        console.log(result.trim());
    }
}

// Function to move the front node from the source list to the destination list
function MoveNode(destRef, sourceRef) {
    if (sourceRef.head === null) return;

    // Take the node from the front of the source
    let newNode = sourceRef.head;
    sourceRef.head = sourceRef.head.next;

    // Move the node to the front of the destRef list
    newNode.next = destRef.head;
    destRef.head = newNode;
}

// Function to split the linked list into two alternating lists
function AlternatingSplit(source, aRef, bRef) {
    let current = source.head;

    // Alternate between adding nodes to 'aRef' and 'bRef'
    while (current !== null) {
        // Move the node to 'aRef'
        MoveNode(aRef, source);

        // If the current node is not null, move the next node to 'bRef'
        if (source.head !== null) {
            MoveNode(bRef, source);
        }

        current = source.head; // Update current to the new head of the source
    }
}

// Create a linked list
let sourceList = new LinkedList();

// Add nodes to the source list
sourceList.append(1);
sourceList.append(2);
sourceList.append(3);
sourceList.append(4);
sourceList.append(5);
sourceList.append(6);

// Create references for the new lists
let aRef = new LinkedList();
let bRef = new LinkedList();

// Print the original list
console.log("Original list:");
sourceList.printList();

// Perform alternating split
AlternatingSplit(sourceList, aRef, bRef);

// Print the split lists
console.log("List A:");
aRef.printList();

console.log("List B:");
bRef.printList();

Output
Original list:
1 2 3 4 5 6
List A:
5 3 1
List B:
6 4 2

Complexity Analysis:

  • Time Complexity: O(n) where n is number of node in the given linked list.
  • Space Complexity: O(1), because we just use some variables.

Please refer complete article on Alternating split of a given Singly Linked List | Set 1 for more details!


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kartik
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