Skip to content
Related Articles

Related Articles

Improve Article

Merge Sort for Linked Lists

  • Difficulty Level : Hard
  • Last Updated : 13 Jul, 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);
 

C++




// C++ code for linked list merged sort
#include <bits/stdc++.h>
using namespace std;
 
/* Link list node */
class Node {
public:
    int data;
    Node* next;
};
 
/* function prototypes */
Node* SortedMerge(Node* a, Node* b);
void FrontBackSplit(Node* source,
                    Node** frontRef, Node** backRef);
 
/* sorts the linked list by changing next pointers (not data) */
void MergeSort(Node** headRef)
{
    Node* head = *headRef;
    Node* a;
    Node* b;
 
    /* Base case -- length 0 or 1 */
    if ((head == NULL) || (head->next == NULL)) {
        return;
    }
 
    /* Split head into 'a' and 'b' sublists */
    FrontBackSplit(head, &a, &b);
 
    /* Recursively sort the sublists */
    MergeSort(&a);
    MergeSort(&b);
 
    /* answer = merge the two sorted lists together */
    *headRef = SortedMerge(a, b);
}
 
/* See https:// www.geeksforgeeks.org/?p=3622 for details of this
function */
Node* SortedMerge(Node* a, Node* b)
{
    Node* result = NULL;
 
    /* Base cases */
    if (a == NULL)
        return (b);
    else if (b == NULL)
        return (a);
 
    /* Pick either a or b, and recur */
    if (a->data <= b->data) {
        result = a;
        result->next = SortedMerge(a->next, b);
    }
    else {
        result = b;
        result->next = SortedMerge(a, b->next);
    }
    return (result);
}
 
/* UTILITY FUNCTIONS */
/* Split the nodes of the given list into front and back halves,
    and return the two lists using the reference parameters.
    If the length is odd, the extra node should go in the front list.
    Uses the fast/slow pointer strategy. */
void FrontBackSplit(Node* source,
                    Node** frontRef, Node** backRef)
{
    Node* fast;
    Node* slow;
    slow = source;
    fast = source->next;
 
    /* Advance 'fast' two nodes, and advance 'slow' one node */
    while (fast != NULL) {
        fast = fast->next;
        if (fast != NULL) {
            slow = slow->next;
            fast = fast->next;
        }
    }
 
    /* 'slow' is before the midpoint in the list, so split it in two
    at that point. */
    *frontRef = source;
    *backRef = slow->next;
    slow->next = NULL;
}
 
/* Function to print nodes in a given linked list */
void printList(Node* node)
{
    while (node != NULL) {
        cout << node->data << " ";
        node = node->next;
    }
}
 
/* Function to insert a node at the beginging of the linked list */
void push(Node** head_ref, int new_data)
{
    /* allocate node */
    Node* new_node = new Node();
 
    /* put in the data */
    new_node->data = new_data;
 
    /* link the old list off the new node */
    new_node->next = (*head_ref);
 
    /* move the head to point to the new node */
    (*head_ref) = new_node;
}
 
/* Driver program to test above functions*/
int main()
{
    /* Start with the empty list */
    Node* res = NULL;
    Node* a = NULL;
 
    /* Let us create a unsorted linked lists to test the functions
Created lists shall be a: 2->3->20->5->10->15 */
    push(&a, 15);
    push(&a, 10);
    push(&a, 5);
    push(&a, 20);
    push(&a, 3);
    push(&a, 2);
 
    /* Sort the above created Linked List */
    MergeSort(&a);
 
    cout << "Sorted Linked List is: \n";
    printList(a);
 
    return 0;
}
 
// This is code is contributed by rathbhupendra


C




// C code for linked list merged sort
#include <stdio.h>
#include <stdlib.h>
 
/* Link list node */
struct Node {
    int data;
    struct Node* next;
};
 
/* function prototypes */
struct Node* SortedMerge(struct Node* a, struct Node* b);
void FrontBackSplit(struct Node* source,
                    struct Node** frontRef, struct Node** backRef);
 
/* sorts the linked list by changing next pointers (not data) */
void MergeSort(struct Node** headRef)
{
    struct Node* head = *headRef;
    struct Node* a;
    struct Node* b;
 
    /* Base case -- length 0 or 1 */
    if ((head == NULL) || (head->next == NULL)) {
        return;
    }
 
    /* Split head into 'a' and 'b' sublists */
    FrontBackSplit(head, &a, &b);
 
    /* Recursively sort the sublists */
    MergeSort(&a);
    MergeSort(&b);
 
    /* answer = merge the two sorted lists together */
    *headRef = SortedMerge(a, b);
}
 
/* See https:// www.geeksforgeeks.org/?p=3622 for details of this
function */
struct Node* SortedMerge(struct Node* a, struct Node* b)
{
    struct Node* result = NULL;
 
    /* Base cases */
    if (a == NULL)
        return (b);
    else if (b == NULL)
        return (a);
 
    /* Pick either a or b, and recur */
    if (a->data <= b->data) {
        result = a;
        result->next = SortedMerge(a->next, b);
    }
    else {
        result = b;
        result->next = SortedMerge(a, b->next);
    }
    return (result);
}
 
/* UTILITY FUNCTIONS */
/* Split the nodes of the given list into front and back halves,
    and return the two lists using the reference parameters.
    If the length is odd, the extra node should go in the front list.
    Uses the fast/slow pointer strategy. */
void FrontBackSplit(struct Node* source,
                    struct Node** frontRef, struct Node** backRef)
{
    struct Node* fast;
    struct Node* slow;
    slow = source;
    fast = source->next;
 
    /* Advance 'fast' two nodes, and advance 'slow' one node */
    while (fast != NULL) {
        fast = fast->next;
        if (fast != NULL) {
            slow = slow->next;
            fast = fast->next;
        }
    }
 
    /* 'slow' is before the midpoint in the list, so split it in two
    at that point. */
    *frontRef = source;
    *backRef = slow->next;
    slow->next = NULL;
}
 
/* Function to print nodes in a given linked list */
void printList(struct Node* node)
{
    while (node != NULL) {
        printf("%d ", node->data);
        node = node->next;
    }
}
 
/* Function to insert a node at the beginging of the linked list */
void push(struct Node** head_ref, int new_data)
{
    /* allocate node */
    struct Node* new_node = (struct Node*)malloc(sizeof(struct Node));
 
    /* put in the data */
    new_node->data = new_data;
 
    /* link the old list off the new node */
    new_node->next = (*head_ref);
 
    /* move the head to point to the new node */
    (*head_ref) = new_node;
}
 
/* Driver program to test above functions*/
int main()
{
    /* Start with the empty list */
    struct Node* res = NULL;
    struct Node* a = NULL;
 
    /* Let us create a unsorted linked lists to test the functions
Created lists shall be a: 2->3->20->5->10->15 */
    push(&a, 15);
    push(&a, 10);
    push(&a, 5);
    push(&a, 20);
    push(&a, 3);
    push(&a, 2);
 
    /* Sort the above created Linked List */
    MergeSort(&a);
 
    printf("Sorted Linked List is: \n");
    printList(a);
 
    getchar();
    return 0;
}


Java




// Java program to illustrate merge sorted
// of linkedList
 
public class linkedList {
    node head = null;
    // node a, b;
    static class node {
        int val;
        node next;
 
        public node(int val)
        {
            this.val = val;
        }
    }
 
    node sortedMerge(node a, node b)
    {
        node 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;
    }
 
    node mergeSort(node h)
    {
        // Base case : if head is null
        if (h == null || h.next == null) {
            return h;
        }
 
        // get the middle of the list
        node middle = getMiddle(h);
        node nextofmiddle = middle.next;
 
        // set the next of middle node to null
        middle.next = null;
 
        // Apply mergeSort on left list
        node left = mergeSort(h);
 
        // Apply mergeSort on right list
        node right = mergeSort(nextofmiddle);
 
        // Merge the left and right lists
        node sortedlist = sortedMerge(left, right);
        return sortedlist;
    }
 
    // Utility function to get the middle of the linked list
    public static node getMiddle(node head)
    {
        if (head == null)
            return head;
 
        node slow = head, fast = head;
 
        while (fast.next != null && fast.next.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        return slow;
    }
 
    void push(int new_data)
    {
        /* allocate node */
        node 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
    void printList(node headref)
    {
        while (headref != null) {
            System.out.print(headref.val + " ");
            headref = headref.next;
        }
    }
 
    public static void main(String[] args)
    {
 
        linkedList li = new linkedList();
        /*
         * Let us create a unsorted linked list to test the functions
         * created. The list shall be a: 2->3->20->5->10->15
         */
        li.push(15);
        li.push(10);
        li.push(5);
        li.push(20);
        li.push(3);
        li.push(2);
 
        // Apply merge Sort
        li.head = li.mergeSort(li.head);
        System.out.print("\n Sorted Linked List is: \n");
        li.printList(li.head);
    }
}
 
// This code is contributed by Rishabh Mahrsee


Python3




# Python3 program to merge sort of linked list
 
# create Node using class Node.
class Node:
    def __init__(self, data):
        self.data = data
        self.next = None
 
class LinkedList:
    def __init__(self):
        self.head = None
     
    # push new value to linked list
    # using append method
    def append(self, new_value):
         
        # Allocate new node
        new_node = Node(new_value)
         
        # if head is None, initialize it to new node
        if self.head is None:
            self.head = new_node
            return
        curr_node = self.head
        while curr_node.next is not None:
            curr_node = curr_node.next
             
        # Append the new node at the end
        # of the linked list
        curr_node.next = new_node
         
    def sortedMerge(self, a, b):
        result = None
         
        # Base cases
        if a == None:
            return b
        if b == None:
            return a
             
        # pick either a or b and recur..
        if a.data <= b.data:
            result = a
            result.next = self.sortedMerge(a.next, b)
        else:
            result = b
            result.next = self.sortedMerge(a, b.next)
        return result
     
    def mergeSort(self, h):
         
        # Base case if head is None
        if h == None or h.next == None:
            return h
 
        # get the middle of the list
        middle = self.getMiddle(h)
        nexttomiddle = middle.next
 
        # set the next of middle node to None
        middle.next = None
 
        # Apply mergeSort on left list
        left = self.mergeSort(h)
         
        # Apply mergeSort on right list
        right = self.mergeSort(nexttomiddle)
 
        # Merge the left and right lists
        sortedlist = self.sortedMerge(left, right)
        return sortedlist
     
    # Utility function to get the middle
    # of the linked list
    def getMiddle(self, head):
        if (head == None):
            return head
 
        slow = head
        fast = head
 
        while (fast.next != None and
               fast.next.next != None):
            slow = slow.next
            fast = fast.next.next
             
        return slow
         
# Utility function to print the linked list
def printList(head):
    if head is None:
        print(' ')
        return
    curr_node = head
    while curr_node:
        print(curr_node.data, end = " ")
        curr_node = curr_node.next
    print(' ')
     
# Driver Code
if __name__ == '__main__':
    li = LinkedList()
     
    # Let us create a unsorted linked list
    # to test the functions created.
    # The list shall be a: 2->3->20->5->10->15
    li.append(15)
    li.append(10)
    li.append(5)
    li.append(20)
    li.append(3)
    li.append(2)
     
    # Apply merge Sort
    li.head = li.mergeSort(li.head)
    print ("Sorted Linked List is:")
    printList(li.head)
 
# This code is contributed by Vikas Chitturi


C#




// C# program to illustrate merge sorted
// of linkedList
using System;
 
public class linkedList {
    node head = null;
 
    // node a, b;
    public class node {
        public int val;
        public node next;
 
        public node(int val)
        {
            this.val = val;
        }
    }
 
    node sortedMerge(node a, node b)
    {
        node 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;
    }
 
    node mergeSort(node h)
    {
        // Base case : if head is null
        if (h == null || h.next == null) {
            return h;
        }
 
        // get the middle of the list
        node middle = getMiddle(h);
        node nextofmiddle = middle.next;
 
        // set the next of middle node to null
        middle.next = null;
 
        // Apply mergeSort on left list
        node left = mergeSort(h);
 
        // Apply mergeSort on right list
        node right = mergeSort(nextofmiddle);
 
        // Merge the left and right lists
        node sortedlist = sortedMerge(left, right);
        return sortedlist;
    }
 
    // Utility function to get the
    // middle of the linked list
    node getMiddle(node h)
    {
        // Base case
        if (h == null)
            return h;
        node fastptr = h.next;
        node slowptr = h;
 
        // Move fastptr by two and slow ptr by one
        // Finally slowptr will point to middle node
        while (fastptr != null) {
            fastptr = fastptr.next;
            if (fastptr != null) {
                slowptr = slowptr.next;
                fastptr = fastptr.next;
            }
        }
        return slowptr;
    }
 
    void push(int new_data)
    {
        /* allocate node */
        node 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
    void printList(node headref)
    {
        while (headref != null) {
            Console.Write(headref.val + " ");
            headref = headref.next;
        }
    }
 
    // Driver code
    public static void Main(String[] args)
    {
 
        linkedList li = new linkedList();
        /*
        * Let us create a unsorted linked list to test the functions
        * created. The list shall be a: 2->3->20->5->10->15
        */
        li.push(15);
        li.push(10);
        li.push(5);
        li.push(20);
        li.push(3);
        li.push(2);
 
        // Apply merge Sort
        li.head = li.mergeSort(li.head);
        Console.Write("\n Sorted Linked List is: \n");
        li.printList(li.head);
    }
}
 
// This code is contributed by Arnab Kundu


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 povar 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("\n Sorted Linked List is: \n");
        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)

Java




// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
// Node Class
class Node {
    int data;
    Node next;
    Node(int key)
    {
        this.data = key;
        next = null;
    }
}
 
class GFG {
   
    // Function to merge sort
    static Node mergeSort(Node head)
    {
        if (head.next == null)
            return head;
 
        Node mid = findMid(head);
        Node head2 = mid.next;
        mid.next = null;
        Node newHead1 = mergeSort(head);
        Node newHead2 = mergeSort(head2);
        Node finalHead = merge(newHead1, newHead2);
 
        return finalHead;
    }
 
    // Function to merge two linked lists
    static Node merge(Node head1, Node head2)
    {
        Node merged = new Node(-1);
        Node 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
    static Node findMid(Node head)
    {
        Node slow = head, fast = head.next;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        return slow;
    }
 
    // Function to print list
    static void printList(Node head)
    {
        while (head != null) {
            System.out.print(head.data + " ");
            head = head.next;
        }
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        Node head = new Node(7);
        Node 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);
        System.out.print("\nSorted Linked List is: \n");
        printList(head);
    }
}


Python3




# Python program for the above approach
 
 
# Node Class
class Node:
     
    def __init__(self,key):
        self.data=key
        self.next=None
 
# Function to merge sort
def mergeSort(head):
    if (head.next == None):
        return head
     
    mid = findMid(head)
    head2 = mid.next
    mid.next = None
    newHead1 = mergeSort(head)
    newHead2 = mergeSort(head2)
    finalHead = merge(newHead1, newHead2)
    return finalHead
 
# Function to merge two linked lists
def merge(head1,head2):
    merged = Node(-1)
     
    temp = merged
    # While head1 is not null and head2
    # is not null
    while (head1 != None and head2 != None):
        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 != None):
        temp.next = head1
        head1 = head1.next
        temp = temp.next
     
    # While head2 is not null
    while (head2 != None):
        temp.next = head2
        head2 = head2.next
        temp = temp.next
     
    return merged.next
 
# Find mid using The Tortoise and The Hare approach
def findMid(head):
    slow = head
    fast = head.next
    while (fast != None and fast.next != None):
        slow = slow.next
        fast = fast.next.next
    return slow
 
# Function to print list
def printList(head):
    while (head != None):
        print(head.data,end=" ")
        head=head.next
 
# Driver Code
head = Node(7)
temp = head
temp.next = Node(10);
temp = temp.next;
temp.next = Node(5);
temp = temp.next;
temp.next = Node(20);
temp = temp.next;
temp.next = Node(3);
temp = temp.next;
temp.next = Node(2);
temp = temp.next;
 
# Apply merge Sort
head = mergeSort(head);
print("\nSorted Linked List is: \n");
 
printList(head);
  
# This code is contributed by avanitrachhadiya2155


C#




// C# program for the above approach
using System;
 
// Node Class
public class Node
{
    public int data;
    public Node next;
 
    public Node(int key)
    {
        this.data = key;
        next = null;
    }
}
 
class GFG{
 
// Function to merge sort
static Node mergeSort(Node head)
{
    if (head.next == null)
        return head;
 
    Node mid = findMid(head);
    Node head2 = mid.next;
    mid.next = null;
    Node newHead1 = mergeSort(head);
    Node newHead2 = mergeSort(head2);
    Node finalHead = merge(newHead1, newHead2);
 
    return finalHead;
}
 
// Function to merge two linked lists
static Node merge(Node head1, Node head2)
{
    Node merged = new Node(-1);
    Node 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
static Node findMid(Node head)
{
    Node slow = head, fast = head.next;
    while (fast != null && fast.next != null)
    {
        slow = slow.next;
        fast = fast.next.next;
    }
    return slow;
}
 
// Function to print list
static void printList(Node head)
{
    while (head != null)
    {
        Console.Write(head.data + " ");
        head = head.next;
    }
}
 
// Driver Code
public static void Main(String[] args)
{
    Node head = new Node(7);
    Node 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);
    Console.Write("\nSorted Linked List is: \n");
    printList(head);
}
}
 
// This code is contributed by umadevi9616


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>


C++




#include<iostream>
using namespace std;
 
//Node structure
struct Node{
    int data;
    Node *next;
};
 
//function to insert in list
void insert(int x,Node **head) {
    if(*head == NULL){
        *head = new Node;
        (*head)->data = x;
        (*head)->next = NULL;
        return;
    }
    Node *temp = new Node;
    temp->data = (*head)->data;
    temp->next = (*head)->next;
    (*head)->data=x;
    (*head)->next=temp;
}
 
//function to print the list
void print(Node *head) {
    Node *temp=head;
    while(temp!=NULL) {
        cout<<temp->data<<" ";
        temp = temp->next;
    }
}
 
Node *merge(Node *firstNode,Node *secondNode) {
    Node *merged = new Node;
    Node *temp = new Node;
    
    //merged is equal to temp so in the end we have the top Node.
    merged = temp;
 
    //while either firstNode or secondNode becomes NULL
    while(firstNode != NULL && secondNode != NULL) {
     
        if(firstNode->data<=secondNode->data) {
            temp->next = firstNode;
            firstNode = firstNode->next;
        }
         
        else {
            temp->next = secondNode;
            secondNode = secondNode->next;
        }
        temp = temp->next;
    }
    
   //any remaining Node in firstNode or secondNode gets inserted in the temp List
    while(firstNode!=NULL) {
        temp->next = firstNode;
        firstNode = firstNode->next;
        temp = temp->next;
    }
 
    while(secondNode!=NULL) {
        temp->next = secondNode;
        secondNode = secondNode->next;
        temp = temp->next;
    }
    //return the head of the sorted list
    return merged->next;
}
 
//function to calculate the middle Element
Node *middle(Node *head) {
    Node *slow = head;
    Node *fast = head->next;
    
    while(slow->next != NULL && (fast!=NULL && fast->next!=NULL)) {
        slow = slow->next;
        fast = fast->next->next;
    }
    return slow;
}
 
//function to sort the given list
Node *sort(Node *head){
     
    if(head->next == NULL) {
        return head;
    }
 
    Node *mid = new Node;
    Node *head2 = new Node;
    mid = middle(head);
    head2 = mid->next;  
    mid->next = NULL;
    //recursive call to sort() hence diving our problem, and then merging the solution
    Node *finalhead = merge(sort(head),sort(head2)); 
    return finalhead;
}
int main(void) {
    Node *head = NULL;
    int n[]={7,10,5,20,3,2};
    for(int i=0;i<6;i++) {
        insert(n[i],&head);   //inserting in the list
    }
  cout<<"\nSorted Linked List is: \n";
    print(sort(head));     //printing the sorted list returned by sort()
    return 0;
}


Output:

Sorted Linked List is: 
2 3 5 7 10 20 

Time Complexity: O(n*log n)

Space Complexity: O(log n)

Sources: 
http://en.wikipedia.org/wiki/Merge_sort 
http://cslibrary.stanford.edu/105/LinkedListProblems.pdf

Please write comments if you find the above code/algorithm incorrect, or find better ways to solve the same problem.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :