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Merge Sort for Linked Lists in JavaScript

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Prerequisite: Merge Sort for Linked Lists Merge sort is often preferred for sorting a linked list. The slow random-access performance of a linked list makes some other algorithms (such as quicksort) perform poorly, and others (such as heapsort) completely impossible.

sorting image

In this post, Merge sort for linked list is implemented using JavaScript.

Examples:

Input : 5 -> 4 -> 3 -> 2 -> 1
Output :1 -> 2 -> 3 -> 4 -> 5

Input : 10 -> 20 -> 3 -> 2 -> 1
Output : 1 -> 2 -> 3 -> 10 -> 20

Implementation:

javascript




<script>
    // Create Node of LinkedList 
    function Node(data) { 
            this.node = data; 
            this.next = null
    
       
    // To initialize a linkedlist 
    function LinkedList(list) { 
        this.head = list || null
    
       
    // Function to insert The new Node into the linkedList 
    LinkedList.prototype.insert = function(data) { 
       
            // Check if the linked list is empty 
            // so insert first node and lead head 
            // points to generic node 
            if (this.head === null
                this.head = new Node(data); 
       
            else
       
                // If linked list is not empty, insert the node 
                // at the end of the linked list 
                let list = this.head; 
                while (list.next) { 
                    list = list.next; 
                
       
                // Now here list pointer points to last 
                // node let’s insert out new node in it 
                list.next = new Node(data) 
            
    
       
    // Function to print linkedList 
    LinkedList.prototype.iterate = function() { 
       
            // First we will check whether out 
            // linked list is empty or node 
            if (this.head === null
                return null
       
            // If linked list is not empty we will 
            // iterate from each Node and prints 
            // it’s value store in “data” property 
       
            let list = this.head; 
       
            // we will iterate until our list variable 
            // contains the “Next” value of the last Node 
            // i.e-> null 
            while (list) { 
                document.write(list.node) 
                if (list.next) 
                    document.write(' -> '
                list = list.next 
            
    
       
    // Function to mergesort a linked list 
    LinkedList.prototype.mergeSort = function(list) { 
       
            if (list.next === null
                return list; 
       
            let count = 0; 
            let countList = list 
            let leftPart = list; 
            let leftPointer = list; 
            let rightPart = null
            let rightPointer = null
       
            // Counting the nodes in the received linkedlist 
            while (countList.next !== null) { 
                count++; 
                countList = countList.next; 
            
       
            // counting the mid of the linked list 
            let mid = Math.floor(count / 2) 
            let count2 = 0; 
       
            // separating the left and right part with 
            // respect to mid node in the linked list 
            while (count2 < mid) { 
                count2++; 
                leftPointer = leftPointer.next; 
            
       
            rightPart = new LinkedList(leftPointer.next); 
            leftPointer.next = null
       
            // Here are two linked list which 
            // contains the left most nodes and right 
            // most nodes of the mid node 
            return this._mergeSort(this.mergeSort(leftPart), 
                                    this.mergeSort(rightPart.head)) 
    
       
    // Merging both lists in a sorted manner 
    LinkedList.prototype._mergeSort = function(left, right) { 
       
            // Create a new empty linked list 
            let result = new LinkedList() 
       
            let resultPointer = result.head; 
            let pointerLeft = left; 
            let pointerRight = right; 
           
                   
            // If true then add left most node value in result, 
            // increment left pointer else do the same in 
            // right linked list. 
            // This loop will be executed until pointer's of 
            // a left node or right node reached null 
            while (pointerLeft && pointerRight) { 
                let tempNode = null
       
                // Check if the right node's value is greater than 
                // left node's value 
                if (pointerLeft.node > pointerRight.node) { 
                    tempNode = pointerRight.node 
                    pointerRight = pointerRight.next; 
                
                else
                    tempNode = pointerLeft.node 
                    pointerLeft = pointerLeft.next; 
                
       
                if (result.head == null) { 
                    result.head = new Node(tempNode) 
                    resultPointer = result.head 
                
                else
                    resultPointer.next = new Node(tempNode) 
                    resultPointer = resultPointer.next 
                
            
       
            // Add the remaining elements in the last of resultant 
            // linked list 
            resultPointer.next = pointerLeft; 
            while (resultPointer.next) 
                resultPointer = resultPointer.next 
       
                resultPointer.next = pointerRight 
       
            // Result is the new sorted linked list 
            return result.head; 
    
       
    // Initialize the object 
    let l = new LinkedList(); 
    l.insert(10) 
    l.insert(20) 
    l.insert(3) 
    l.insert(2) 
    l.insert(1) 
    // Print the linked list 
    l.iterate() 
       
    // Sort the linked list 
    l.head = LinkedList.prototype.mergeSort(l.head) 
       
    document.write('<br> After sorting : '); 
       
    // Print the sorted linked list 
    l.iterate() 
</script>


Output

10 -> 20 -> 3 -> 2 -> 1
After sorting : 1 -> 2 -> 3 -> 10 -> 20 

Complexity Analysis:

  • Time Complexity: O(n*log(n))
  • Auxiliary Space: O(n)


Last Updated : 30 Mar, 2023
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