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# Merge two sorted linked lists

• Difficulty Level : Medium
• Last Updated : 25 Jan, 2023

AuxiliaryGiven two sorted linked lists consisting of N and M nodes respectively. The task is to merge both of the lists (in place) and return the head of the merged list.

Examples:

Input: a: 5->10->15, b: 2->3->20
Output: 2->3->5->10->15->20

Input: a: 1->1, b: 2->4
Output: 1->1->2->4 ## Brute Force Way:

The Approach:

In this Approach we use a vector of (n+m) size where n and m are length respective linked list and then store all the element in vector and then we sort the vector and make new linked list it will be our answer.

## C++

 `#include ``using` `namespace` `std;`` ` `/* Link list Node */``struct` `Node {``    ``int` `key;``    ``struct` `Node* next;``};` `Node* newNode(``int` `key){``    ``Node* temp = ``new` `Node;``    ``temp->key = key;``    ``temp->next = NULL;``    ``return` `temp;``}``int` `main() {``    ``/* Let us create two sorted linked lists to test``       ``the above functions. Created lists shall be``         ``a: 5->10->15->40``         ``b: 2->3->20  */``    ``Node* a = newNode(5);``    ``a->next = newNode(10);``    ``a->next->next = newNode(15);``    ``a->next->next->next = newNode(40);`` ` `    ``Node* b = newNode(2);``    ``b->next = newNode(3);``    ``b->next->next = newNode(20);``   ``vector<``int``>v;``    ``while``(a!=NULL){``     ``v.push_back(a->key);``     ``a=a->next;``    ``}``     ``while``(b!=NULL){``     ``v.push_back(b->key);``     ``b=b->next;``    ``}``     ``sort(v.begin(),v.end());``     ``Node* result= newNode(-1);``     ``Node* temp=result;``     ``for``(``int` `i=0;inext=newNode(v[i]);``      ``result=result->next;``     ``}``     ``temp=temp->next;``     ``cout<<``"Resultant Merge Linked List Is :"``<key<<``" "``;``       ``temp=temp->next;``     ``}``    ``return` `0;``}`

## Javascript

 `// JavaScript program for the above approach``// Link List node``class Node{``    ``constructor(key){``        ``this``.key = key;``        ``this``.next = ``null``;``    ``}``}` `function` `newNode(key){``    ``let temp = ``new` `Node(key);``    ``return` `temp;``}` `// driver code``// let us create two sorted linked lists to test the above``// function. Created lists shall be``// a: 5->10->15->40``// b: 2->3->20` `let a = newNode(5);``a.next = newNode(10);``a.next.next = newNode(15);``a.next.next.next = newNode(40);` `let b = newNode(2);``b.next = newNode(3);``b.next.next = newNode(20);` `let v = [];``while``(a != ``null``){``    ``v.push(a.key);``    ``a = a.next;``}` `while``(b != ``null``){``    ``v.push(b.key);``    ``b = b.next;``}` `v.sort(``function``(a, b){``return` `a - b});``let result = newNode(-1);``let temp = result;``for``(let i = 0; i

Output

```Resultant Merge Linked List Is :
2 3 5 10 15 20 40 ```

Complexity Analysis:

Time Complexity:O((n+m)*(n+m)log(n+m)),(n+m)for traversing linked lists and (n+m)log(n+m) for sorting vector.
Auxiliary Space: O(n+m),for vector.

## Merge two sorted linked lists using Dummy Nodes:

The idea is to use a temporary dummy node as the start of the result list. The pointer Tail always points to the last node in the result list, so appending new nodes is easy.

Follow the below illustration for a better understanding:

Illustration: Follow the steps below to solve the problem:

• First, make a dummy node for the new merged linked list
• Now make two pointers, one will point to list1 and another will point to list2.
• Now traverse the lists till one of them gets exhausted.
• If the value of the node pointing to either list is smaller than another pointer, add that node to our merged list and increment that pointer.

Below is the implementation of the above approach:

## C++

 `/* C++ program to merge two sorted linked lists */``#include ``using` `namespace` `std;` `/* Link list node */``class` `Node {``public``:``    ``int` `data;``    ``Node* next;``};` `/* pull of the front node of``the source and put it in dest */``void` `MoveNode(Node** destRef, Node** sourceRef);` `/* Takes two lists sorted in increasing``order, and splices their nodes together``to make one big sorted list which``is returned. */``Node* SortedMerge(Node* a, Node* b)``{``    ``/* a dummy first node to hang the result on */``    ``Node dummy;` `    ``/* tail points to the last result node */``    ``Node* tail = &dummy;` `    ``/* so tail->next is the place to``    ``add new nodes to the result. */``    ``dummy.next = NULL;``    ``while` `(1) {``        ``if` `(a == NULL) {``            ``/* if either list runs out, use the``            ``other list */``            ``tail->next = b;``            ``break``;``        ``}``        ``else` `if` `(b == NULL) {``            ``tail->next = a;``            ``break``;``        ``}``        ``if` `(a->data <= b->data)``            ``MoveNode(&(tail->next), &a);``        ``else``            ``MoveNode(&(tail->next), &b);` `        ``tail = tail->next;``    ``}``    ``return` `(dummy.next);``}` `/* UTILITY FUNCTIONS */``/* MoveNode() function takes the``node from the front of the source,``and move it to the front of the dest.``It is an error to call this with the``source list empty.` `Before calling MoveNode():``source == {1, 2, 3}``dest == {1, 2, 3}` `After calling MoveNode():``source == {2, 3}``dest == {1, 1, 2, 3} */``void` `MoveNode(Node** destRef, Node** sourceRef)``{``    ``/* the front source node */``    ``Node* newNode = *sourceRef;``    ``assert``(newNode != NULL);` `    ``/* Advance the source pointer */``    ``*sourceRef = newNode->next;` `    ``/* Link the old dest of the new node */``    ``newNode->next = *destRef;` `    ``/* Move dest to point to the new node */``    ``*destRef = newNode;``}` `/* Function to insert a node at``the beginning 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 of the new node */``    ``new_node->next = (*head_ref);` `    ``/* move the head to point to the new node */``    ``(*head_ref) = new_node;``}` `/* Function to print nodes in a given linked list */``void` `printList(Node* node)``{``    ``while` `(node != NULL) {``        ``cout << node->data << ``" "``;``        ``node = node->next;``    ``}``}` `/* Driver code*/``int` `main()``{``    ``/* Start with the empty list */``    ``Node* res = NULL;``    ``Node* a = NULL;``    ``Node* b = NULL;` `    ``/* Let us create two sorted linked lists``    ``to test the functions``    ``Created lists, a: 5->10->15, b: 2->3->20 */``    ``push(&a, 15);``    ``push(&a, 10);``    ``push(&a, 5);` `    ``push(&b, 20);``    ``push(&b, 3);``    ``push(&b, 2);` `    ``/* Remove duplicates from linked list */``    ``res = SortedMerge(a, b);` `    ``cout << ``"Merged Linked List is: \n"``;``    ``printList(res);` `    ``return` `0;``}` `// This code is contributed by rathbhupendra`

## C

 `/* C program to merge two sorted linked lists */``#include ``#include ``#include ` `/* Link list node */``struct` `Node {``    ``int` `data;``    ``struct` `Node* next;``};` `/* pull of the front node of the source and put it in dest`` ``*/``void` `MoveNode(``struct` `Node** destRef,``              ``struct` `Node** sourceRef);` `/* Takes two lists sorted in increasing order, and splices``   ``their nodes together to make one big sorted list which``   ``is returned.  */``struct` `Node* SortedMerge(``struct` `Node* a, ``struct` `Node* b)``{``    ``/* a dummy first node to hang the result on */``    ``struct` `Node dummy;` `    ``/* tail points to the last result node  */``    ``struct` `Node* tail = &dummy;` `    ``/* so tail->next is the place to add new nodes``      ``to the result. */``    ``dummy.next = NULL;``    ``while` `(1) {``        ``if` `(a == NULL) {``            ``/* if either list runs out, use the``               ``other list */``            ``tail->next = b;``            ``break``;``        ``}``        ``else` `if` `(b == NULL) {``            ``tail->next = a;``            ``break``;``        ``}``        ``if` `(a->data <= b->data)``            ``MoveNode(&(tail->next), &a);``        ``else``            ``MoveNode(&(tail->next), &b);` `        ``tail = tail->next;``    ``}``    ``return` `(dummy.next);``}` `/* UTILITY FUNCTIONS */``/* MoveNode() function takes the node from the front of the``   ``source, and move it to the front of the dest.``   ``It is an error to call this with the source list empty.` `   ``Before calling MoveNode():``   ``source == {1, 2, 3}``   ``dest == {1, 2, 3}` `   ``After calling MoveNode():``   ``source == {2, 3}``   ``dest == {1, 1, 2, 3} */``void` `MoveNode(``struct` `Node** destRef,``              ``struct` `Node** sourceRef)``{``    ``/* the front source node  */``    ``struct` `Node* newNode = *sourceRef;``    ``assert``(newNode != NULL);` `    ``/* Advance the source pointer */``    ``*sourceRef = newNode->next;` `    ``/* Link the old dest of the new node */``    ``newNode->next = *destRef;` `    ``/* Move dest to point to the new node */``    ``*destRef = newNode;``}` `/* Function to insert a node at the beginning 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 of the new node */``    ``new_node->next = (*head_ref);` `    ``/* move the head to point to the new node */``    ``(*head_ref) = new_node;``}` `/* Function to print nodes in a given linked list */``void` `printList(``struct` `Node* node)``{``    ``while` `(node != NULL) {``        ``printf``(``"%d "``, node->data);``        ``node = node->next;``    ``}``}` `/* Driver program to test above functions*/``int` `main()``{``    ``/* Start with the empty list */``    ``struct` `Node* res = NULL;``    ``struct` `Node* a = NULL;``    ``struct` `Node* b = NULL;` `    ``/* Let us create two sorted linked lists to test``      ``the functions``       ``Created lists, a: 5->10->15,  b: 2->3->20 */``    ``push(&a, 15);``    ``push(&a, 10);``    ``push(&a, 5);` `    ``push(&b, 20);``    ``push(&b, 3);``    ``push(&b, 2);` `    ``/* Remove duplicates from linked list */``    ``res = SortedMerge(a, b);` `    ``printf``(``"Merged Linked List is: \n"``);``    ``printList(res);` `    ``return` `0;``}`

## Java

 `/* Java program to merge two``   ``sorted linked lists */``import` `java.util.*;` `/* Link list node */``class` `Node {``    ``int` `data;``    ``Node next;``    ``Node(``int` `d)``    ``{``        ``data = d;``        ``next = ``null``;``    ``}``}` `class` `MergeLists {``    ``Node head;` `    ``/* Method to insert a node at``       ``the end of the linked list */``    ``public` `void` `addToTheLast(Node node)``    ``{``        ``if` `(head == ``null``) {``            ``head = node;``        ``}``        ``else` `{``            ``Node temp = head;``            ``while` `(temp.next != ``null``)``                ``temp = temp.next;``            ``temp.next = node;``        ``}``    ``}` `    ``/* Method to print linked list */``    ``void` `printList()``    ``{``        ``Node temp = head;``        ``while` `(temp != ``null``) {``            ``System.out.print(temp.data + ``" "``);``            ``temp = temp.next;``        ``}``        ``System.out.println();``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String args[])``    ``{``        ``/* Let us create two sorted linked``           ``lists to test the methods``           ``Created lists:``               ``llist1: 5->10->15,``               ``llist2: 2->3->20``        ``*/``        ``MergeLists llist1 = ``new` `MergeLists();``        ``MergeLists llist2 = ``new` `MergeLists();` `        ``// Node head1 = new Node(5);``        ``llist1.addToTheLast(``new` `Node(``5``));``        ``llist1.addToTheLast(``new` `Node(``10``));``        ``llist1.addToTheLast(``new` `Node(``15``));` `        ``// Node head2 = new Node(2);``        ``llist2.addToTheLast(``new` `Node(``2``));``        ``llist2.addToTheLast(``new` `Node(``3``));``        ``llist2.addToTheLast(``new` `Node(``20``));` `        ``llist1.head = ``new` `Gfg().sortedMerge(llist1.head,``                                            ``llist2.head);``        ``System.out.println(``"Merged Linked List is:"``);``        ``llist1.printList();``    ``}``}` `class` `Gfg {``    ``/* Takes two lists sorted in``    ``increasing order, and splices``    ``their nodes together to make``    ``one big sorted list which is``    ``returned. */``    ``Node sortedMerge(Node headA, Node headB)``    ``{` `        ``/* a dummy first node to``           ``hang the result on */``        ``Node dummyNode = ``new` `Node(``0``);` `        ``/* tail points to the``        ``last result node */``        ``Node tail = dummyNode;``        ``while` `(``true``) {` `            ``/* if either list runs out,``            ``use the other list */``            ``if` `(headA == ``null``) {``                ``tail.next = headB;``                ``break``;``            ``}``            ``if` `(headB == ``null``) {``                ``tail.next = headA;``                ``break``;``            ``}` `            ``/* Compare the data of the two``            ``lists whichever lists' data is``            ``smaller, append it into tail and``            ``advance the head to the next Node``            ``*/``            ``if` `(headA.data <= headB.data) {``                ``tail.next = headA;``                ``headA = headA.next;``            ``}``            ``else` `{``                ``tail.next = headB;``                ``headB = headB.next;``            ``}` `            ``/* Advance the tail */``            ``tail = tail.next;``        ``}``        ``return` `dummyNode.next;``    ``}``}` `// This code is contributed``// by Shubhaw Kumar`

## Python3

 `""" Python program to merge two``sorted linked lists """`  `# Linked List Node``class` `Node:``    ``def` `__init__(``self``, data):``        ``self``.data ``=` `data``        ``self``.``next` `=` `None`  `# Create & Handle List operations``class` `LinkedList:``    ``def` `__init__(``self``):``        ``self``.head ``=` `None` `    ``# Method to display the list``    ``def` `printList(``self``):``        ``temp ``=` `self``.head``        ``while` `temp:``            ``print``(temp.data, end``=``" "``)``            ``temp ``=` `temp.``next` `    ``# Method to add element to list``    ``def` `addToList(``self``, newData):``        ``newNode ``=` `Node(newData)``        ``if` `self``.head ``is` `None``:``            ``self``.head ``=` `newNode``            ``return` `        ``last ``=` `self``.head``        ``while` `last.``next``:``            ``last ``=` `last.``next` `        ``last.``next` `=` `newNode`  `# Function to merge the lists``# Takes two lists which are sorted``# joins them to get a single sorted list``def` `mergeLists(headA, headB):` `    ``# A dummy node to store the result``    ``dummyNode ``=` `Node(``0``)` `    ``# Tail stores the last node``    ``tail ``=` `dummyNode``    ``while` `True``:` `        ``# If any of the list gets completely empty``        ``# directly join all the elements of the other list``        ``if` `headA ``is` `None``:``            ``tail.``next` `=` `headB``            ``break``        ``if` `headB ``is` `None``:``            ``tail.``next` `=` `headA``            ``break` `        ``# Compare the data of the lists and whichever is smaller is``        ``# appended to the last of the merged list and the head is changed``        ``if` `headA.data <``=` `headB.data:``            ``tail.``next` `=` `headA``            ``headA ``=` `headA.``next``        ``else``:``            ``tail.``next` `=` `headB``            ``headB ``=` `headB.``next` `        ``# Advance the tail``        ``tail ``=` `tail.``next` `    ``# Returns the head of the merged list``    ``return` `dummyNode.``next`  `# Create 2 lists``listA ``=` `LinkedList()``listB ``=` `LinkedList()` `# Add elements to the list in sorted order``listA.addToList(``5``)``listA.addToList(``10``)``listA.addToList(``15``)` `listB.addToList(``2``)``listB.addToList(``3``)``listB.addToList(``20``)` `# Call the merge function``listA.head ``=` `mergeLists(listA.head, listB.head)` `# Display merged list``print``(``"Merged Linked List is:"``)``listA.printList()` `""" This code is contributed``by Debidutta Rath """`

## C#

 `/* C# program to merge two``sorted linked lists */``using` `System;` `/* Link list node */``public` `class` `Node {``    ``public` `int` `data;``    ``public` `Node next;``    ``public` `Node(``int` `d)``    ``{``        ``data = d;``        ``next = ``null``;``    ``}``}` `public` `class` `MergeLists {``    ``Node head;` `    ``/* Method to insert a node at``    ``the end of the linked list */``    ``public` `void` `addToTheLast(Node node)``    ``{``        ``if` `(head == ``null``) {``            ``head = node;``        ``}``        ``else` `{``            ``Node temp = head;``            ``while` `(temp.next != ``null``)``                ``temp = temp.next;``            ``temp.next = node;``        ``}``    ``}` `    ``/* Method to print linked list */``    ``void` `printList()``    ``{``        ``Node temp = head;``        ``while` `(temp != ``null``) {``            ``Console.Write(temp.data + ``" "``);``            ``temp = temp.next;``        ``}``        ``Console.WriteLine();``    ``}` `    ``// Driver Code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``/* Let us create two sorted linked``        ``lists to test the methods``        ``Created lists:``               ``llist1: 5->10->15,``            ``llist2: 2->3->20``        ``*/``        ``MergeLists llist1 = ``new` `MergeLists();``        ``MergeLists llist2 = ``new` `MergeLists();` `        ``// Node head1 = new Node(5);``        ``llist1.addToTheLast(``new` `Node(5));``        ``llist1.addToTheLast(``new` `Node(10));``        ``llist1.addToTheLast(``new` `Node(15));` `        ``// Node head2 = new Node(2);``        ``llist2.addToTheLast(``new` `Node(2));``        ``llist2.addToTheLast(``new` `Node(3));``        ``llist2.addToTheLast(``new` `Node(20));` `        ``llist1.head = ``new` `Gfg().sortedMerge(llist1.head,``                                            ``llist2.head);``        ``Console.WriteLine(``"Merged Linked List is:"``);``        ``llist1.printList();``    ``}``}` `public` `class` `Gfg {``    ``/* Takes two lists sorted in``    ``increasing order, and splices``    ``their nodes together to make``    ``one big sorted list which is``    ``returned. */``    ``public` `Node sortedMerge(Node headA, Node headB)``    ``{` `        ``/* a dummy first node to``        ``hang the result on */``        ``Node dummyNode = ``new` `Node(0);` `        ``/* tail points to the``        ``last result node */``        ``Node tail = dummyNode;``        ``while` `(``true``) {` `            ``/* if either list runs out,``            ``use the other list */``            ``if` `(headA == ``null``) {``                ``tail.next = headB;``                ``break``;``            ``}``            ``if` `(headB == ``null``) {``                ``tail.next = headA;``                ``break``;``            ``}` `            ``/* Compare the data of the two``            ``lists whichever lists' data is``            ``smaller, append it into tail and``            ``advance the head to the next Node``            ``*/``            ``if` `(headA.data <= headB.data) {``                ``tail.next = headA;``                ``headA = headA.next;``            ``}``            ``else` `{``                ``tail.next = headB;``                ``headB = headB.next;``            ``}` `            ``/* Advance the tail */``            ``tail = tail.next;``        ``}``        ``return` `dummyNode.next;``    ``}``}` `// This code is contributed 29AjayKumar`

## Javascript

 ``

Output

```Merged Linked List is:
2 3 5 10 15 20 ```

Time Complexity: O(N + M), where N and M are the size of list1 and list2 respectively
Auxiliary Space: O(1)

## Merge two sorted linked lists using Recursion:

The idea is to move ahead with node in the recursion whose node value is lesser. When any of the node reach the end then append the rest of the linked List.

Follow the steps below to solve the problem:

• Make a function where two pointers pointing to the linked list will be passed.
• Now, check which value is less from both the current nodes
• The one with less value makes a recursion call by moving ahead with that pointer and simultaneously append that recursion call with the node
• Also put two base cases to check whether one of the linked lists will reach the NULL, then append the rest of the linked list.

Below is the implementation of the above approach.

## C++

 `/* C++ program to merge two sorted linked lists */``#include ``using` `namespace` `std;` `/* Link list node */``class` `Node {``public``:``    ``int` `data;``    ``Node* next;``};` `/* Takes two lists sorted in increasing``order, and splices their nodes together``to make one big sorted list which``is returned. */``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);``}` `/* Function to insert a node at``the beginning 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 of the new node */``    ``new_node->next = (*head_ref);` `    ``/* move the head to point to the new node */``    ``(*head_ref) = new_node;``}` `/* Function to print nodes in a given linked list */``void` `printList(Node* node)``{``    ``while` `(node != NULL) {``        ``cout << node->data << ``" "``;``        ``node = node->next;``    ``}``}` `/* Driver code*/``int` `main()``{``    ``/* Start with the empty list */``    ``Node* res = NULL;``    ``Node* a = NULL;``    ``Node* b = NULL;` `    ``/* Let us create two sorted linked lists``    ``to test the functions``    ``Created lists, a: 5->10->15, b: 2->3->20 */``    ``push(&a, 15);``    ``push(&a, 10);``    ``push(&a, 5);` `    ``push(&b, 20);``    ``push(&b, 3);``    ``push(&b, 2);` `    ``/* Remove duplicates from linked list */``    ``res = SortedMerge(a, b);` `    ``cout << ``"Merged Linked List is: \n"``;``    ``printList(res);` `    ``return` `0;``}` `// This code is contributed by rathbhupendra`

## C

 `/* C program to merge two sorted linked lists */``#include ``#include ``#include ` `/* Link list node */``struct` `Node {``    ``int` `data;``    ``struct` `Node* next;``};` `/* pull off the front node of the source and put it in dest`` ``*/``void` `MoveNode(``struct` `Node** destRef,``              ``struct` `Node** sourceRef);` `/* Takes two lists sorted in increasing order, and splices``   ``their nodes together to make one big sorted list which``   ``is returned.  */``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 */``/* MoveNode() function takes the node from the front of the``   ``source, and move it to the front of the dest.``   ``It is an error to call this with the source list empty.` `   ``Before calling MoveNode():``   ``source == {1, 2, 3}``   ``dest == {1, 2, 3}` `   ``After calling MoveNode():``   ``source == {2, 3}``   ``dest == {1, 1, 2, 3} */``void` `MoveNode(``struct` `Node** destRef,``              ``struct` `Node** sourceRef)``{``    ``/* the front source node  */``    ``struct` `Node* newNode = *sourceRef;``    ``assert``(newNode != NULL);` `    ``/* Advance the source pointer */``    ``*sourceRef = newNode->next;` `    ``/* Link the old dest off the new node */``    ``newNode->next = *destRef;` `    ``/* Move dest to point to the new node */``    ``*destRef = newNode;``}` `/* Function to insert a node at the beginning 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 of the new node */``    ``new_node->next = (*head_ref);` `    ``/* move the head to point to the new node */``    ``(*head_ref) = new_node;``}` `/* Function to print nodes in a given linked list */``void` `printList(``struct` `Node* node)``{``    ``while` `(node != NULL) {``        ``printf``(``"%d "``, node->data);``        ``node = node->next;``    ``}``}` `/* Driver program to test above functions*/``int` `main()``{``    ``/* Start with the empty list */``    ``struct` `Node* res = NULL;``    ``struct` `Node* a = NULL;``    ``struct` `Node* b = NULL;` `    ``/* Let us create two sorted linked lists to test``      ``the functions``       ``Created lists, a: 5->10->15,  b: 2->3->20 */``    ``push(&a, 15);``    ``push(&a, 10);``    ``push(&a, 5);` `    ``push(&b, 20);``    ``push(&b, 3);``    ``push(&b, 2);` `    ``/* Remove duplicates from linked list */``    ``res = SortedMerge(a, b);` `    ``printf``(``"Merged Linked List is: \n"``);``    ``printList(res);` `    ``return` `0;``}`

## Java

 `/* Java program to merge two``   ``sorted linked lists */``import` `java.util.*;` `/* Link list node */``class` `Node {``    ``int` `data;``    ``Node next;``    ``Node(``int` `d)``    ``{``        ``data = d;``        ``next = ``null``;``    ``}``}` `class` `MergeLists {``    ``Node head;` `    ``/* Method to insert a node at``       ``the end of the linked list */``    ``public` `void` `addToTheLast(Node node)``    ``{``        ``if` `(head == ``null``) {``            ``head = node;``        ``}``        ``else` `{``            ``Node temp = head;``            ``while` `(temp.next != ``null``)``                ``temp = temp.next;``            ``temp.next = node;``        ``}``    ``}` `    ``/* Method to print linked list */``    ``void` `printList()``    ``{``        ``System.out.println(``"Merged Linked List is:"``);``        ``Node temp = head;``        ``while` `(temp != ``null``) {``            ``System.out.print(temp.data + ``" "``);``            ``temp = temp.next;``        ``}``        ``System.out.println();``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String args[])``    ``{``        ``/* Let us create two sorted linked``           ``lists to test the methods``           ``Created lists:``               ``llist1: 5->10->15,``               ``llist2: 2->3->20``        ``*/``        ``MergeLists llist1 = ``new` `MergeLists();``        ``MergeLists llist2 = ``new` `MergeLists();` `        ``// Node head1 = new Node(5);``        ``llist1.addToTheLast(``new` `Node(``5``));``        ``llist1.addToTheLast(``new` `Node(``10``));``        ``llist1.addToTheLast(``new` `Node(``15``));` `        ``// Node head2 = new Node(2);``        ``llist2.addToTheLast(``new` `Node(``2``));``        ``llist2.addToTheLast(``new` `Node(``3``));``        ``llist2.addToTheLast(``new` `Node(``20``));` `        ``llist1.head = ``new` `Gfg().sortedMerge(llist1.head,``                                            ``llist2.head);``        ``llist1.printList();``    ``}``}` `class` `Gfg {``    ``/* Takes two lists sorted in``    ``increasing order, and splices``    ``their nodes together to make``    ``one big sorted list which is``    ``returned. */``    ``public` `Node sortedMerge(Node A, Node B)``    ``{` `        ``if` `(A == ``null``)``            ``return` `B;``        ``if` `(B == ``null``)``            ``return` `A;` `        ``if` `(A.data < B.data) {``            ``A.next = sortedMerge(A.next, B);``            ``return` `A;``        ``}``        ``else` `{``            ``B.next = sortedMerge(A, B.next);``            ``return` `B;``        ``}``    ``}``}` `// This code is contributed by Tuhin Das`

## Python3

 `# Python3 program merge two sorted linked``# in third linked list using recursive.` `# Node class`  `class` `Node:``    ``def` `__init__(``self``, data):``        ``self``.data ``=` `data``        ``self``.``next` `=` `None`  `# Constructor to initialize the node object``class` `LinkedList:` `    ``# Function to initialize head``    ``def` `__init__(``self``):``        ``self``.head ``=` `None` `    ``# Method to print linked list``    ``def` `printList(``self``):``        ``temp ``=` `self``.head` `        ``while` `temp:``            ``print``(temp.data, end``=``" "``)``            ``temp ``=` `temp.``next` `    ``# Function to add of node at the end.``    ``def` `append(``self``, new_data):``        ``new_node ``=` `Node(new_data)` `        ``if` `self``.head ``is` `None``:``            ``self``.head ``=` `new_node``            ``return``        ``last ``=` `self``.head` `        ``while` `last.``next``:``            ``last ``=` `last.``next``        ``last.``next` `=` `new_node`  `# Function to merge two sorted linked list.``def` `mergeLists(head1, head2):` `    ``# create a temp node NULL``    ``temp ``=` `None` `    ``# List1 is empty then return List2``    ``if` `head1 ``is` `None``:``        ``return` `head2` `    ``# if List2 is empty then return List1``    ``if` `head2 ``is` `None``:``        ``return` `head1` `    ``# If List1's data is smaller or``    ``# equal to List2's data``    ``if` `head1.data <``=` `head2.data:` `        ``# assign temp to List1's data``        ``temp ``=` `head1` `        ``# Again check List1's data is smaller or equal List2's``        ``# data and call mergeLists function.``        ``temp.``next` `=` `mergeLists(head1.``next``, head2)` `    ``else``:``        ``# If List2's data is greater than or equal List1's``        ``# data assign temp to head2``        ``temp ``=` `head2` `        ``# Again check List2's data is greater or equal List's``        ``# data and call mergeLists function.``        ``temp.``next` `=` `mergeLists(head1, head2.``next``)` `    ``# return the temp list.``    ``return` `temp`  `# Driver Function``if` `__name__ ``=``=` `'__main__'``:` `    ``# Create linked list :``    ``list1 ``=` `LinkedList()``    ``list1.append(``5``)``    ``list1.append(``10``)``    ``list1.append(``15``)` `    ``# Create linked list 2 :``    ``list2 ``=` `LinkedList()``    ``list2.append(``2``)``    ``list2.append(``3``)``    ``list2.append(``20``)` `    ``# Create linked list 3``    ``list3 ``=` `LinkedList()` `    ``# Merging linked list 1 and linked list 2``    ``# in linked list 3``    ``list3.head ``=` `mergeLists(list1.head, list2.head)` `    ``print``(``"Merged Linked List is:"``)``    ``list3.printList()`  `# This code is contributed by 'Shriaknt13'.`

## C#

 `/* C# program to merge two``sorted linked lists */``using` `System;` `/* Link list node */``public` `class` `Node {``    ``public` `int` `data;``    ``public` `Node next;``    ``public` `Node(``int` `d)``    ``{``        ``data = d;``        ``next = ``null``;``    ``}``}` `public` `class` `MergeLists {``    ``Node head;` `    ``/* Method to insert a node at``    ``the end of the linked list */``    ``public` `void` `addToTheLast(Node node)``    ``{``        ``if` `(head == ``null``) {``            ``head = node;``        ``}``        ``else` `{``            ``Node temp = head;``            ``while` `(temp.next != ``null``)``                ``temp = temp.next;``            ``temp.next = node;``        ``}``    ``}` `    ``/* Method to print linked list */``    ``void` `printList()``    ``{``        ``Console.WriteLine(``"Merged Linked List is:"``);``        ``Node temp = head;``        ``while` `(temp != ``null``) {``            ``Console.Write(temp.data + ``" "``);``            ``temp = temp.next;``        ``}``        ``Console.WriteLine();``    ``}` `    ``// Driver Code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``/* Let us create two sorted linked``        ``lists to test the methods``        ``Created lists:``               ``llist1: 5->10->15,``            ``llist2: 2->3->20``        ``*/``        ``MergeLists llist1 = ``new` `MergeLists();``        ``MergeLists llist2 = ``new` `MergeLists();` `        ``// Node head1 = new Node(5);``        ``llist1.addToTheLast(``new` `Node(5));``        ``llist1.addToTheLast(``new` `Node(10));``        ``llist1.addToTheLast(``new` `Node(15));` `        ``// Node head2 = new Node(2);``        ``llist2.addToTheLast(``new` `Node(2));``        ``llist2.addToTheLast(``new` `Node(3));``        ``llist2.addToTheLast(``new` `Node(20));` `        ``llist1.head = ``new` `Gfg().sortedMerge(llist1.head,``                                            ``llist2.head);``        ``llist1.printList();``    ``}``}` `public` `class` `Gfg {``    ``/* Takes two lists sorted in``    ``increasing order, and splices``    ``their nodes together to make``    ``one big sorted list which is``    ``returned. */``    ``public` `Node sortedMerge(Node A, Node B)``    ``{` `        ``// Base cases``        ``if` `(A == ``null``)``            ``return` `B;``        ``if` `(B == ``null``)``            ``return` `A;` `        ``// Pick either a or b, and recur``        ``if` `(A.data < B.data) {``            ``A.next = sortedMerge(A.next, B);``            ``return` `A;``        ``}``        ``else` `{``            ``B.next = sortedMerge(A, B.next);``            ``return` `B;``        ``}``    ``}``}`

## Javascript

 ``

Output

```Merged Linked List is:
2 3 5 10 15 20 ```

Time Complexity: O(M + N), Where M and N are the size of the list1 and list2 respectively.
Auxiliary Space: O(M+N), Function call stack space

## Merge two sorted linked lists by Reversing the Lists:

This idea involves first reversing both the given lists and after reversing, traversing both the lists till the end and then comparing the nodes of both the lists and inserting the node with a larger value at the beginning of the result list. And in this way, we will get the resulting list in increasing order.

Follow the steps below to solve the problem:

• Initialize result list as empty: head = NULL.
• Let ‘a’ and ‘b’ be the heads of the first and second lists respectively.
• Reverse both lists.
• While (a != NULL and b != NULL)
• Find the larger of two (Current ‘a’ and ‘b’)
• Insert the larger value of the node at the front of result list.
• Move ahead in the list with the larger node.
• If ‘b’ becomes NULL before ‘a’, insert all nodes of ‘a’ into result list at the beginning.
• If ‘a’ becomes NULL before ‘b’, insert all nodes of ‘b’ into result list at the beginning.

Below is the implementation of the above solution.

## C++

 `/*Given two sorted linked lists consisting of N and M nodes``respectively. The task is to merge both of the list``(in-place) and return head of the merged list.*/``#include ``using` `namespace` `std;` `/* Link list Node */``struct` `Node {``    ``int` `key;``    ``struct` `Node* next;``};` `// Function to reverse a given Linked List using Recursion``Node* reverseList(Node* head)``{``    ``if` `(head->next == NULL)``        ``return` `head;``    ``Node* rest = reverseList(head->next);``    ``head->next->next = head;``    ``head->next = NULL;``    ``return` `rest;``}` `// Given two non-empty linked lists 'a' and 'b'``Node* sortedMerge(Node* a, Node* b)``{``    ``// Reverse Linked List 'a'``    ``a = reverseList(a);``    ``// Reverse Linked List 'b'``    ``b = reverseList(b);``    ``// Initialize head of resultant list``    ``Node* head = NULL;``    ``Node* temp;``    ``// Traverse both lists while both of them``    ``// have nodes.``    ``while` `(a != NULL && b != NULL) {``        ``// If a's current value is greater than or equal to``        ``// b's current value.``        ``if` `(a->key >= b->key) {``            ``// Store next of current Node in first list``            ``temp = a->next;``            ``// Add 'a' at the front of resultant list``            ``a->next = head;``            ``// Make 'a' - head of the result list``            ``head = a;``            ``// Move ahead in first list``            ``a = temp;``        ``}` `        ``// If b's value is greater. Below steps are similar``        ``// to above (Only 'a' is replaced with 'b')``        ``else` `{``            ``temp = b->next;``            ``b->next = head;``            ``head = b;``            ``b = temp;``        ``}``    ``}` `    ``// If second list reached end, but first list has``    ``// nodes. Add remaining nodes of first list at the``    ``// beginning of result list``    ``while` `(a != NULL) {``        ``temp = a->next;``        ``a->next = head;``        ``head = a;``        ``a = temp;``    ``}` `    ``// If first list reached end, but second list has``    ``// nodes. Add remaining nodes of second list at the``    ``// beginning of result list``    ``while` `(b != NULL) {``        ``temp = b->next;``        ``b->next = head;``        ``head = b;``        ``b = temp;``    ``}``    ``// Return the head of the result list``    ``return` `head;``}` `/* Function to print Nodes in a given linked list */``void` `printList(``struct` `Node* Node)``{``    ``while` `(Node != NULL) {``        ``cout << Node->key << ``" "``;``        ``Node = Node->next;``    ``}``}` `/* Utility function to create a new node with``   ``given key */``Node* newNode(``int` `key)``{``    ``Node* temp = ``new` `Node;``    ``temp->key = key;``    ``temp->next = NULL;``    ``return` `temp;``}` `/* Driver program to test above functions*/``int` `main()``{``    ``/* Start with the empty list */``    ``struct` `Node* res = NULL;` `    ``/* Let us create two sorted linked lists to test``       ``the above functions. Created lists shall be``         ``a: 5->10->15->40``         ``b: 2->3->20  */``    ``Node* a = newNode(5);``    ``a->next = newNode(10);``    ``a->next->next = newNode(15);``    ``a->next->next->next = newNode(40);` `    ``Node* b = newNode(2);``    ``b->next = newNode(3);``    ``b->next->next = newNode(20);` `    ``/* merge 2 sorted Linked Lists */``    ``res = sortedMerge(a, b);` `    ``cout << ``"Merged Linked List is:"` `<< endl;``    ``printList(res);` `    ``return` `0;``}` `// This code is contributed by Aditya Kumar (adityakumar129)`

## C

 `/*Given two sorted linked lists consisting of N and M nodes``respectively. The task is to merge both of the list``(in-place) and return head of the merged list.*/``#include ``#include ` `/* Link list Node */``typedef` `struct` `Node {``    ``int` `key;``    ``struct` `Node* next;``} Node;` `// Function to reverse a given Linked List using Recursion``Node* reverseList(Node* head)``{``    ``if` `(head->next == NULL)``        ``return` `head;``    ``Node* rest = reverseList(head->next);``    ``head->next->next = head;``    ``head->next = NULL;``    ``return` `rest;``}` `// Given two non-empty linked lists 'a' and 'b'``Node* sortedMerge(Node* a, Node* b)``{``    ``// Reverse Linked List 'a'``    ``a = reverseList(a);``    ``// Reverse Linked List 'b'``    ``b = reverseList(b);``    ``// Initialize head of resultant list``    ``Node* head = NULL;``    ``Node* temp;``    ``// Traverse both lists while both of them``    ``// have nodes.``    ``while` `(a != NULL && b != NULL) {``        ``// If a's current value is greater than or equal to``        ``// b's current value.``        ``if` `(a->key >= b->key) {``            ``// Store next of current Node in first list``            ``temp = a->next;``            ``// Add 'a' at the front of resultant list``            ``a->next = head;``            ``// Make 'a' - head of the result list``            ``head = a;``            ``// Move ahead in first list``            ``a = temp;``        ``}` `        ``// If b's value is greater. Below steps are similar``        ``// to above (Only 'a' is replaced with 'b')``        ``else` `{``            ``temp = b->next;``            ``b->next = head;``            ``head = b;``            ``b = temp;``        ``}``    ``}` `    ``// If second list reached end, but first list has``    ``// nodes. Add remaining nodes of first list at the``    ``// beginning of result list``    ``while` `(a != NULL) {``        ``temp = a->next;``        ``a->next = head;``        ``head = a;``        ``a = temp;``    ``}` `    ``// If first list reached end, but second list has``    ``// nodes. Add remaining nodes of second list at the``    ``// beginning of result list``    ``while` `(b != NULL) {``        ``temp = b->next;``        ``b->next = head;``        ``head = b;``        ``b = temp;``    ``}``    ``// Return the head of the result list``    ``return` `head;``}` `/* Function to print Nodes in a given linked list */``void` `printList(``struct` `Node* Node)``{``    ``while` `(Node != NULL) {``        ``printf``(``"%d  "``, Node->key);``        ``Node = Node->next;``    ``}``}` `/* Utility function to create a new node with``   ``given key */``Node* newNode(``int` `key)``{``    ``Node* temp = (Node*)``malloc``(``sizeof``(Node));``    ``temp->key = key;``    ``temp->next = NULL;``    ``return` `temp;``}` `/* Driver program to test above functions*/``int` `main()``{``    ``/* Start with the empty list */``    ``struct` `Node* res = NULL;``    ``/* Let us create two sorted linked lists to test``       ``the above functions. Created lists shall be``         ``a: 5->10->15->40``         ``b: 2->3->20  */``    ``Node* a = newNode(5);``    ``a->next = newNode(10);``    ``a->next->next = newNode(15);``    ``a->next->next->next = newNode(40);` `    ``Node* b = newNode(2);``    ``b->next = newNode(3);``    ``b->next->next = newNode(20);` `    ``/* merge 2 sorted Linked Lists */``    ``res = sortedMerge(a, b);` `    ``printf``(``"Merged Linked List is: \n"``);``    ``printList(res);` `    ``return` `0;``}` `// This code is contributed by Aditya Kumar (adityakumar129)`

## Java

 `/*Given two sorted linked lists consisting of N and M nodes``respectively. The task is to merge both of the list``(in-place) and return head of the merged list.*/` `import` `java.util.*;` `class` `GFG {` `    ``/* Link list Node */``    ``static` `class` `Node {``        ``int` `key;``        ``Node next;``    ``};` `    ``// Function to reverse a given Linked List using``    ``// Recursion``    ``static` `Node reverseList(Node head)``    ``{` `        ``if` `(head.next == ``null``)``            ``return` `head;` `        ``Node rest = reverseList(head.next);``        ``head.next.next = head;``        ``head.next = ``null``;` `        ``return` `rest;``    ``}` `    ``// Given two non-empty linked lists 'a' and 'b'``    ``static` `Node sortedMerge(Node a, Node b)``    ``{` `        ``// Reverse Linked List 'a'``        ``a = reverseList(a);` `        ``// Reverse Linked List 'b'``        ``b = reverseList(b);` `        ``// Initialize head of resultant list``        ``Node head = ``null``;` `        ``Node temp;` `        ``// Traverse both lists while both of them``        ``// have nodes.``        ``while` `(a != ``null` `&& b != ``null``) {` `            ``// If a's current value is greater than or equal``            ``// to b's current value.``            ``if` `(a.key >= b.key) {` `                ``// Store next of current Node in first list``                ``temp = a.next;` `                ``// Add 'a' at the front of resultant list``                ``a.next = head;` `                ``// Make 'a' - head of the result list``                ``head = a;` `                ``// Move ahead in first list``                ``a = temp;``            ``}` `            ``// If b's value is greater. Below steps are``            ``// similar to above (Only 'a' is replaced with``            ``// 'b')``            ``else` `{` `                ``temp = b.next;``                ``b.next = head;``                ``head = b;``                ``b = temp;``            ``}``        ``}` `        ``// If second list reached end, but first list has``        ``// nodes. Add remaining nodes of first list at the``        ``// beginning of result list``        ``while` `(a != ``null``) {` `            ``temp = a.next;``            ``a.next = head;``            ``head = a;``            ``a = temp;``        ``}` `        ``// If first list reached end, but second list has``        ``// nodes. Add remaining nodes of second list at the``        ``// beginning of result list``        ``while` `(b != ``null``) {` `            ``temp = b.next;``            ``b.next = head;``            ``head = b;``            ``b = temp;``        ``}` `        ``// Return the head of the result list``        ``return` `head;``    ``}` `    ``/* Function to print Nodes in a given linked list */``    ``static` `void` `printList(Node Node)``    ``{``        ``while` `(Node != ``null``) {``            ``System.out.print(Node.key + ``" "``);``            ``Node = Node.next;``        ``}``    ``}` `    ``/* Utility function to create a new node with``       ``given key */``    ``static` `Node newNode(``int` `key)``    ``{``        ``Node temp = ``new` `Node();``        ``temp.key = key;``        ``temp.next = ``null``;``        ``return` `temp;``    ``}` `    ``/* Driver program to test above functions*/``    ``public` `static` `void` `main(String[] args)``    ``{``        ``/* Start with the empty list */``        ``Node res = ``null``;` `        ``/* Let us create two sorted linked lists to test``           ``the above functions. Created lists shall be``             ``a: 5.10.15.40``             ``b: 2.3.20  */``        ``Node a = newNode(``5``);``        ``a.next = newNode(``10``);``        ``a.next.next = newNode(``15``);``        ``a.next.next.next = newNode(``40``);` `        ``Node b = newNode(``2``);``        ``b.next = newNode(``3``);``        ``b.next.next = newNode(``20``);` `        ``/* merge 2 sorted Linked Lists */``        ``res = sortedMerge(a, b);` `        ``System.out.println(``"Merged Linked List is:"``);``        ``printList(res);``    ``}``}` `// This code is contributed by umadevi9616`

## Python3

 `# Given two sorted linked lists consisting of N and M nodes``# respectively. The task is to merge both of the list``# (in-place) and return head of the merged list.` `# Link list Node``class` `Node:``    ``def` `__init__(``self``, key):``        ``self``.key ``=` `key``        ``self``.``next` `=` `None` `# Function to reverse a given Linked List using``# Recursion``def` `reverse_list(head):``    ``if` `head.``next` `is` `None``:``        ``return` `head``    ``rest ``=` `reverse_list(head.``next``)``    ``head.``next``.``next` `=` `head``    ``head.``next` `=` `None``    ``return` `rest` `# Given two non-empty linked lists 'a' and 'b'``def` `sorted_merge(a, b):``    ``# Reverse Linked List 'a'``    ``a ``=` `reverse_list(a)` `    ``# Reverse Linked List 'b'``    ``b ``=` `reverse_list(b)` `    ``# Initialize head of resultant list``    ``head ``=` `None` `    ``# Traverse both lists while both of them``    ``# have nodes.``    ``while` `a ``is` `not` `None` `and` `b ``is` `not` `None``:``        ``# If a's current value is greater than or equal``        ``# to b's current value.``        ``if` `a.key >``=` `b.key:``            ``# Store next of current Node in first list``            ``temp ``=` `a.``next``            ``# Add 'a' at the front of resultant list``            ``a.``next` `=` `head``            ``# Make 'a' - head of the result list``            ``head ``=` `a``            ``# Move ahead in first list``            ``a ``=` `temp``        ``# If b's value is greater. Below steps are``        ``# similar to above (Only 'a' is replaced with``        ``# 'b')``        ``else``:``            ``temp ``=` `b.``next``            ``b.``next` `=` `head``            ``head ``=` `b``            ``b ``=` `temp` `    ``# If second list reached end, but first list has``    ``# nodes. Add remaining nodes of first list at the``    ``# beginning of result list``    ``while` `a ``is` `not` `None``:``        ``temp ``=` `a.``next``        ``a.``next` `=` `head``        ``head ``=` `a``        ``a ``=` `temp` `    ``# If first list reached end, but second list has``    ``# nodes. Add remaining nodes of second list at the``    ``# beginning of result list``    ``while` `b ``is` `not` `None``:``        ``temp ``=` `b.``next``        ``b.``next` `=` `head``        ``head ``=` `b``        ``b ``=` `temp` `    ``# Return the head of the result list``    ``return` `head` `# Function to print Nodes in a given linked list``def` `print_list(node):``    ``while` `node ``is` `not` `None``:``        ``print``(node.key, end``=``" "``)``        ``node ``=` `node.``next``    ``print``()` `# Utility function to create a new node with``# given key``def` `new_node(key):``    ``temp ``=` `Node(key)``    ``return` `temp` `# Test` `# Start with the empty list``res ``=` `None` `# Let us create two sorted linked lists to test``# the above functions. Created lists shall be``#   a: 5.10.15.40``#   b: 2.3.20``a ``=` `new_node(``5``)``a.``next` `=` `new_node(``10``)``a.``next``.``next` `=` `new_node(``15``)``a.``next``.``next``.``next` `=` `new_node(``40``)` `b ``=` `new_node(``2``)``b.``next` `=` `new_node(``3``)``b.``next``.``next` `=` `new_node(``20``)` `# merge 2 sorted linked lists``res ``=` `sorted_merge(a, b)` `print``(``"Merged Linked List is:"``)``print_list(res)` `# This code is contributed by lokesh`

## C#

 `/*Given two sorted linked lists consisting of N and M nodes``respectively. The task is to merge both of the list``(in-place) and return head of the merged list.*/` `using` `System;``using` `System.Collections.Generic;` `public` `class` `GFG {` `    ``/* Link list Node */``    ``public` `class` `Node {``        ``public` `int` `key;``        ``public` `Node next;``    ``};` `    ``// Function to reverse a given Linked List using``    ``// Recursion``    ``static` `Node reverseList(Node head)``    ``{` `        ``if` `(head.next == ``null``)``            ``return` `head;` `        ``Node rest = reverseList(head.next);``        ``head.next.next = head;``        ``head.next = ``null``;` `        ``return` `rest;``    ``}` `    ``// Given two non-empty linked lists 'a' and 'b'``    ``static` `Node sortedMerge(Node a, Node b)``    ``{` `        ``// Reverse Linked List 'a'``        ``a = reverseList(a);` `        ``// Reverse Linked List 'b'``        ``b = reverseList(b);` `        ``// Initialize head of resultant list``        ``Node head = ``null``;` `        ``Node temp;` `        ``// Traverse both lists while both of them``        ``// have nodes.``        ``while` `(a != ``null` `&& b != ``null``) {` `            ``// If a's current value is greater than or equal``            ``// to b's current value.``            ``if` `(a.key >= b.key) {` `                ``// Store next of current Node in first list``                ``temp = a.next;` `                ``// Add 'a' at the front of resultant list``                ``a.next = head;` `                ``// Make 'a' - head of the result list``                ``head = a;` `                ``// Move ahead in first list``                ``a = temp;``            ``}` `            ``// If b's value is greater. Below steps are``            ``// similar to above (Only 'a' is replaced with``            ``// 'b')``            ``else` `{` `                ``temp = b.next;``                ``b.next = head;``                ``head = b;``                ``b = temp;``            ``}``        ``}` `        ``// If second list reached end, but first list has``        ``// nodes. Add remaining nodes of first list at the``        ``// beginning of result list``        ``while` `(a != ``null``) {` `            ``temp = a.next;``            ``a.next = head;``            ``head = a;``            ``a = temp;``        ``}` `        ``// If first list reached end, but second list has``        ``// nodes. Add remaining nodes of second list at the``        ``// beginning of result list``        ``while` `(b != ``null``) {` `            ``temp = b.next;``            ``b.next = head;``            ``head = b;``            ``b = temp;``        ``}` `        ``// Return the head of the result list``        ``return` `head;``    ``}` `    ``/* Function to print Nodes in a given linked list */``    ``static` `void` `printList(Node Node)``    ``{``        ``while` `(Node != ``null``) {``            ``Console.Write(Node.key + ``" "``);``            ``Node = Node.next;``        ``}``    ``}` `    ``/* Utility function to create a new node with``       ``given key */``    ``static` `Node newNode(``int` `key)``    ``{``        ``Node temp = ``new` `Node();``        ``temp.key = key;``        ``temp.next = ``null``;``        ``return` `temp;``    ``}` `    ``/* Driver program to test above functions*/``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``/* Start with the empty list */``        ``Node res = ``null``;` `        ``/* Let us create two sorted linked lists to test``           ``the above functions. Created lists shall be``             ``a: 5.10.15.40``             ``b: 2.3.20  */``        ``Node a = newNode(5);``        ``a.next = newNode(10);``        ``a.next.next = newNode(15);``        ``a.next.next.next = newNode(40);` `        ``Node b = newNode(2);``        ``b.next = newNode(3);``        ``b.next.next = newNode(20);` `        ``/* merge 2 sorted Linked Lists */``        ``res = sortedMerge(a, b);` `        ``Console.WriteLine(``"Merged Linked List is:"``);``        ``printList(res);``    ``}``}` `// This code is contributed by umadevi9616`

## Javascript

 `/*Given two sorted linked lists consisting of N and M nodes``respectively. The task is to merge both of the list``(in-place) and return head of the merged list.*/` `class Node {``    ``constructor(key) {``        ``this``.key = key;``        ``this``.next = ``null``;``    ``}``}` `// Function to reverse a given linked list using recursion``function` `reverseList(head) {``    ``if` `(!head.next) {``        ``return` `head;``    ``}``    ``const rest = reverseList(head.next);``    ``head.next.next = head;``    ``head.next = ``null``;``    ``return` `rest;``}` `// Given two non-empty linked lists 'a' and 'b'``function` `sortedMerge(a, b) {``    ``// Reverse linked list 'a'``    ``a = reverseList(a);``    ``// Reverse linked list 'b'``    ``b = reverseList(b);` `    ``// Initialize head of resultant list``    ``let head = ``null``;``    ``let temp;` `    ``// Traverse both lists while both of them have nodes``    ``while` `(a && b) {``        ``// If a's current value is greater than or equal to b's current value``        ``if` `(a.key >= b.key) {``            ``// Store next of current node in first list``            ``temp = a.next;``            ``// Add 'a' at the front of resultant list``            ``a.next = head;``            ``// Make 'a' the head of the result list``            ``head = a;``            ``// Move ahead in first list``            ``a = temp;``        ``}``        ``// If b's value is greater. Below steps are similar to above (only 'a' is replaced with 'b')``        ``else` `{``            ``temp = b.next;``            ``b.next = head;``            ``head = b;``            ``b = temp;``        ``}``    ``}` `    ``// If second list reached end, but first list has nodes.``    ``// Add remaining nodes of first list at the beginning of result list``    ``while` `(a) {``        ``temp = a.next;``        ``a.next = head;``        ``head = a;``        ``a = temp;``    ``}` `    ``// If first list reached end, but second list has nodes.``    ``// Add remaining nodes of second list at the beginning of result list``    ``while` `(b) {``        ``temp = b.next;``        ``b.next = head;``        ``head = b;``        ``b = temp;``    ``}` `    ``// Return the head of the result list``    ``return` `head;``}` `// Function to print nodes in a given linked list``function` `printList(node) {``    ``let result = ``""``;``    ``while` `(node) {``        ``result += node.key + ``" "``;``        ``node = node.next;``    ``}``    ``console.log(result);``}` `// Start with the empty list``let res = ``null``;` `// Create two sorted linked lists to test the above functions``// List a: 5 -> 10 -> 15 -> 40``// List b: 2 -> 3 -> 20``const a = ``new` `Node(5);``a.next = ``new` `Node(10);``a.next.next = ``new` `Node(15);``a.next.next.next = ``new` `Node(40);` `const b = ``new` `Node(2);``b.next = ``new` `Node(3);``b.next.next = ``new` `Node(20);` `/* merge 2 sorted Linked Lists */``res = sortedMerge(a, b);` `console.log(``"Merged Linked List is:"` `+ ``"
"``);``printList(res);` `// This code is contributed by lokeshmvs21.`

Output

```Merged Linked List is:
2 3 5 10 15 20 40 ```

Time Complexity: O(M+N) where M and N are the lengths of the two lists to be merged.
Auxiliary Space: O(M+N)

This method is contributed by Mehul Mathur(mathurmehul01)

Please refer below post for simpler implementations :
Merge two sorted lists (in-place)
Please write comments if you find the above code/algorithm incorrect, or find better ways to solve the same problem.

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