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Javascript Program For Merge Sort Of Linked Lists

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  • Last Updated : 19 Dec, 2021

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

Let the head be the first node of the linked list to be sorted and headRef be the pointer to head. Note that we need a reference to head in MergeSort() as the below implementation changes next links to sort the linked lists (not data at the nodes), so the head node has to be changed if the data at the original head is not the smallest value in the linked list. 

MergeSort(headRef)
1) If the head is NULL or there is only one element in the Linked List 
    then return.
2) Else divide the linked list into two halves.  
      FrontBackSplit(head, &a, &b); /* a and b are two halves */
3) Sort the two halves a and b.
      MergeSort(a);
      MergeSort(b);
4) Merge the sorted a and b (using SortedMerge() discussed here) 
   and update the head pointer using headRef.
     *headRef = SortedMerge(a, b);
 

Javascript




<script>
 
// Javascript program to
// illustrate merge sorted
// of linkedList
 
 
    var head = null;
 
    // node a, b;
     class node {
            constructor(val) {
                this.val = val;
                this.next = null;
            }
        }
 
    function sortedMerge( a,  b)
    {
        var result = null;
        /* Base cases */
        if (a == null)
            return b;
        if (b == null)
            return a;
 
        /* Pick either a or b, and recur */
        if (a.val <= b.val) {
            result = a;
            result.next = sortedMerge(a.next, b);
        } else {
            result = b;
            result.next = sortedMerge(a, b.next);
        }
        return result;
    }
 
    function mergeSort( h) {
        // Base case : if head is null
        if (h == null || h.next == null) {
            return h;
        }
 
        // get the middle of the list
        var middle = getMiddle(h);
        var nextofmiddle = middle.next;
 
        // set the next of middle node to null
        middle.next = null;
 
        // Apply mergeSort on left list
        var left = mergeSort(h);
 
        // Apply mergeSort on right list
        var right = mergeSort(nextofmiddle);
 
        // Merge the left and right lists
        var sortedlist = sortedMerge(left, right);
        return sortedlist;
    }
 
    // Utility function to get the middle
    // of the linked list
    function getMiddle( head) {
        if (head == null)
            return head;
 
        var slow = head, fast = head;
 
        while (fast.next != null && fast.next.next != null)
        {
            slow = slow.next;
            fast = fast.next.next;
        }
        return slow;
    }
 
    function push(new_data) {
        /* allocate node */
        var new_node = new node(new_data);
 
        /* link the old list off the new node */
        new_node.next = head;
 
        /* move the head to point to the new node */
        head = new_node;
    }
 
    // Utility function to print the linked list
    function printList( headref) {
        while (headref != null) {
            document.write(headref.val + " ");
            headref = headref.next;
        }
    }
 
     
  
        /*
         Let us create a unsorted linked
         list to test the functions
         created. The list shall be
         a: 2->3->20->5->10->15
         */
        push(15);
        push(10);
        push(5);
        push(20);
        push(3);
        push(2);
 
        // Apply merge Sort
        head = mergeSort(head);
        document.write("
 Sorted Linked List is:
");
        printList(head);
 
// This code contributed by umadevi9616
 
</script>


Output: 

Sorted Linked List is: 
2 3 5 10 15 20

 

Time Complexity: O(n*log n)

Space Complexity: O(n*log n)

Approach 2: This approach is simpler and uses log n space.

mergeSort():

  1. If the size of the linked list is 1 then return the head
  2. Find mid using The Tortoise and The Hare Approach
  3. Store the next of mid in head2 i.e. the right sub-linked list.
  4. Now Make the next midpoint null.
  5. Recursively call mergeSort() on both left and right sub-linked list and store the new head of the left and right linked list.
  6. Call merge() given the arguments new heads of left and right sub-linked lists and store the final head returned after merging.
  7. Return the final head of the merged linkedlist.

merge(head1, head2):

  1. Take a pointer say merged to store the merged list in it and store a dummy node in it.
  2. Take a pointer temp and assign merge to it.
  3. If the data of head1 is less than the data of head2, then, store head1 in next of temp & move head1 to the next of head1.
  4. Else store head2 in next of temp & move head2 to the next of head2.
  5. Move temp to the next of temp.
  6. Repeat steps 3, 4 & 5 until head1 is not equal to null and head2 is not equal to null.
  7. Now add any remaining nodes of the first or the second linked list to the merged linked list.
  8. Return the next of merged(that will ignore the dummy and return the head of the final merged linked list)

Javascript




<script>
 
// JavaScript program for the above approach
 
// Node Class
class Node {
    constructor(val) {
        this.data = val;
        this.next = null;
    }
}
    // Function to merge sort
    function mergeSort(head) {
        if (head.next == null)
            return head;
 
var mid = findMid(head);
var head2 = mid.next;
        mid.next = null;
var newHead1 = mergeSort(head);
var newHead2 = mergeSort(head2);
var finalHead = merge(newHead1, newHead2);
 
        return finalHead;
    }
 
    // Function to merge two linked lists
    function merge(head1,  head2) {
var merged = new Node(-1);
var temp = merged;
 
        // While head1 is not null and head2
        // is not null
        while (head1 != null && head2 != null) {
            if (head1.data < head2.data) {
                temp.next = head1;
                head1 = head1.next;
            } else {
                temp.next = head2;
                head2 = head2.next;
            }
            temp = temp.next;
        }
 
        // While head1 is not null
        while (head1 != null) {
            temp.next = head1;
            head1 = head1.next;
            temp = temp.next;
        }
 
        // While head2 is not null
        while (head2 != null) {
            temp.next = head2;
            head2 = head2.next;
            temp = temp.next;
        }
        return merged.next;
    }
 
    // Find mid using The Tortoise and The Hare approach
    function findMid(head) {
var slow = head, fast = head.next;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        return slow;
    }
 
    // Function to print list
    function printList(head) {
        while (head != null) {
            document.write(head.data + " ");
            head = head.next;
        }
    }
 
    // Driver Code
     
var head = new Node(7);
var temp = head;
        temp.next = new Node(10);
        temp = temp.next;
        temp.next = new Node(5);
        temp = temp.next;
        temp.next = new Node(20);
        temp = temp.next;
        temp.next = new Node(3);
        temp = temp.next;
        temp.next = new Node(2);
        temp = temp.next;
 
        // Apply merge Sort
        head = mergeSort(head);
        document.write("Sorted Linked List is: <br/>");
        printList(head);
 
// This code contributed by gauravrajput1
 
</script>


Output:

Sorted Linked List is: 
2 3 5 7 10 20 

Time Complexity: O(n*log n)

Space Complexity: O(log n)

Please refer complete article on Merge Sort for Linked Lists for more details!
 


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