JavaScript Program to Swap Two Elements in a Linked List

We are given a linked list and the task is to swap two elements in a linked list without disturbing their links. There are multiple ways to swap. It can be done by swapping the elements inside the nodes, and by swapping the complete nodes. 

Example:

Input: list = 10 -> 11 -> 12 -> 13 -> 14 -> 15
ele1 = 11
ele2 = 14
Output: list = 10 -> 14 -> 12 -> 13 -> 11 -> 15

Below are the ways by which we can swap two elements in a linked list in JavaScript.

Using Recursion

The approach is to use the recursive nature of JavaScript functions to traverse the linked list and swap the elements.

Algorithm:

• Define a Node class with attributes data and next to represent each element of the linked list.
• Define a LinkedList class with a constructor to initialize the head of the linked list.
• Implement a method insert(data) to insert a new node at the end of the linked list.
• Implement a recursive method swapUtil(current, x, y):
  • If the current node is null, return.
  • If the data of the current node equals x, set its data to y.
  • If the data of the current node equals y, set its data to x.
  • Recursively call swapUtil with the next node as the current node.
• Implement a method swap(x, y):
  • If x and y are the same, return.
  • Call swapUtil with the head of the linked list, x, and y.
• Implement a method printList() to print the elements of the linked list

Example: This example shows the use of the above-explained approach.

Javascript

class Node {
    constructor(data) {
        this.data = data;
        this.next = null;
    }
}

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

    // Function to insert a new node at 
    // the end of the linked list
    insert(data) {
        const newNode = new Node(data);
        if (!this.head) {
            this.head = newNode;
        } else {
            let current = this.head;
            while (current.next) {
                current = current.next;
            }
            current.next = newNode;
        }
    }

    // Function to swap two elements in the 
    // linked list using recursion
    swapUtil(current, x, y) {
        if (!current) return;

        if (current.data === x) {
            current.data = y;
        } else if (current.data === y) {
            current.data = x;
        }

        this.swapUtil(current.next, x, y);
    }

    swap(x, y) {
        if (x === y) return;

        this.swapUtil(this.head, x, y);
    }

    // Function to print the linked list
    printList() {
        let current = this.head;
        while (current) {
            console.log(current.data);
            current = current.next;
        }
    }
}

// Example usage:
const linkedList = new LinkedList();
linkedList.insert(10);
linkedList.insert(11);
linkedList.insert(12);
linkedList.insert(13);
linkedList.insert(14);
linkedList.insert(15);
console.log("Before swapping:");
linkedList.printList();

linkedList.swap(12, 14);
console.log("After swapping:");
linkedList.printList();

Output

Before swapping:
10
11
12
13
14
15
After swapping:
10
11
14
13
12
15

Time Complexity: O(n) where n is the number of nodes in the linked list.
Space Complexity: O(1) because it uses a constant amount of extra space regardless of the size of the input linked list.

Using Iteration

The approach here is to define traditional linked list structure with a Node class representing each element and a LinkedList class to manage the list. It iteratively traverses the list to find the nodes containing the given data values and adjusts their pointers to swap positions.

Algorithm:

1. Define a Node class to represent each element in the linked list with a data field and a next pointer.
2. Define a LinkedList class with a head pointer initially set to null.
3. Implement an append method to add elements to the end of the linked list.
4. Implement a printList method to print the elements of the linked list.
5. Implement a swapNodes method that takes two data values to swap.
6. Traverse the linked list to find the nodes containing the given data values.
7. Swap the positions of the nodes by adjusting their next pointers and the pointers of their previous nodes.
8. Print the updated list.

Example: This example shows the use of the above-explained approach.

Javascript

// Node class to define the structure of 
// each node in the linked list
class Node {
    constructor(data) {
        this.data = data;
        this.next = null;
    }
}

// Linked List class to define the linked 
// list and methods to operate on it
class LinkedList {
    constructor() {
        this.head = null;
    }

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

    // Function to print the linked list
    printList() {
        let current = this.head;
        while (current) {
            console.log(current.data);
            current = current.next;
        }
    }

    // Function to swap two elements in the linked list
    swapNodes(data1, data2) {
        // If both data are the same, nothing to do
        if (data1 === data2) {
            return;
        }

        let prevX = null,
            currX = this.head;
        while (currX && currX.data !== data1) {
            prevX = currX;
            currX = currX.next;
        }

        let prevY = null,
            currY = this.head;
        while (currY && currY.data !== data2) {
            prevY = currY;
            currY = currY.next;
        }

        // If either data is not present, nothing to do
        if (!currX || !currY) {
            return;
        }

        // If data1 is not the head of the linked list
        if (prevX) {
            prevX.next = currY;
        } else {
            this.head = currY;
        }

        // If data2 is not the head of the linked list
        if (prevY) {
            prevY.next = currX;
        } else {
            this.head = currX;
        }

        // Swap next pointers
        let temp = currX.next;
        currX.next = currY.next;
        currY.next = temp;
    }
}

// Example usage
let ll = new LinkedList();
ll.append(10);
ll.append(11);
ll.append(12);
ll.append(13);
ll.append(14);
ll.append(15);

console.log("Linked List Before Swapping:");
ll.printList();

ll.swapNodes(11, 14);

console.log("\nLinked List After Swapping:");
ll.printList();

Output

Linked List Before Swapping:
10
11
12
13
14
15

Linked List After Swapping:
10
14
12
13
11
15

Time Complexity: O(n) for traversing the list to find the nodes to swap and O(1) for swapping them.
Space Complexity: O(1)