# Merge two sorted linked lists

• Difficulty Level : Medium
• Last Updated : 24 Jun, 2022

Write a SortedMerge() function that takes two lists, each of which is sorted in increasing order, and merges the two together into one list which is in increasing order. SortedMerge() should return the new list. The new list should be made by splicing together the nodes of the first two lists.

For example if the first linked list a is 5->10->15 and the other linked list b is 2->3->20, then SortedMerge() should return a pointer to the head node of the merged list 2->3->5->10->15->20.

There are many cases to deal with: either ‘a’ or ‘b’ may be empty, during processing either ‘a’ or ‘b’ may run out first, and finally, there’s the problem of starting the result list empty, and building it up while going through ‘a’ and ‘b’.

Method 1 (Using Dummy Nodes)
The strategy here uses 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.

The dummy node gives the tail something to point to initially when the result list is empty. This dummy node is efficient, since it is only temporary, and it is allocated in the stack. The loop proceeds, removing one node from either ‘a’ or ‘b’, and adding it to the tail. When

We are done, the result is in dummy.next. The below image is a dry run of the above approach: 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 off 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 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(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;  ` `}  ` ` `  `/* Function to print nodes in a given linked list */` `void` `printList(Node *node)  ` `{  ` `    ``while` `(node!=NULL)  ` `    ``{  ` `        ``cout<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) ` `{ ` `    ``/* 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 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 off 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); ` `    ``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);  ` `            ``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```

Method 2 (Using Local References)
This solution is structurally very similar to the above, but it avoids using a dummy node. Instead, it maintains a struct node** pointer, lastPtrRef, that always points to the last pointer of the result list. This solves the same case that the dummy node did — dealing with the result list when it is empty. If you are trying to build up a list at its tail, either the dummy node or the struct node** “reference” strategy can be used (see Section 1 for details).

## C++14

 `Node* SortedMerge(Node* a, Node* b) ` `{ ` `    ``Node* result = NULL; ` ` `  `    ``/* point to the last result pointer */` `    ``Node** lastPtrRef = &result; ` ` `  `    ``while` `(1) { ` `        ``if` `(a == NULL) { ` `            ``*lastPtrRef = b; ` `            ``break``; ` `        ``} ` `        ``else` `if` `(b == NULL) { ` `            ``*lastPtrRef = a; ` `            ``break``; ` `        ``} ` `        ``if` `(a->data <= b->data) ` `            ``MoveNode(lastPtrRef, &a); ` `        ``else` `            ``MoveNode(lastPtrRef, &b); ` ` `  `        ``/* tricky: advance to point to the next ".next" ` `         ``* field */` `        ``lastPtrRef = &((*lastPtrRef)->next); ` `    ``} ` `    ``return` `(result); ` `} ` ` `  `// This code is contributed by Aditya Kumar (adityakumar129)`

## C

 `struct` `Node* SortedMerge(``struct` `Node* a, ``struct` `Node* b) ` `{ ` `    ``struct` `Node* result = NULL; ` ` `  `    ``/* point to the last result pointer */` `    ``struct` `Node** lastPtrRef = &result; ` ` `  `    ``while` `(1) { ` `        ``if` `(a == NULL) { ` `            ``*lastPtrRef = b; ` `            ``break``; ` `        ``} ` `        ``else` `if` `(b == NULL) { ` `            ``*lastPtrRef = a; ` `            ``break``; ` `        ``} ` `        ``if` `(a->data <= b->data) ` `            ``MoveNode(lastPtrRef, &a); ` `        ``else` `            ``MoveNode(lastPtrRef, &b); ` ` `  `        ``/* tricky: advance to point to the next ".next" ` `         ``* field */` `        ``lastPtrRef = &((*lastPtrRef)->next); ` `    ``} ` `    ``return` `(result); ` `} ` ` `  `// This code is contributed by Aditya Kumar (adityakumar129)`

## Java

 `Node SortedMerge(Node a, Node b) ` `{ ` `    ``Node result = ``null``; ` ` `  `    ``/* point to the last result pointer */` `    ``Node lastPtrRef = result; ` ` `  `    ``while` `(``1``) { ` `        ``if` `(a == ``null``) { ` `            ``lastPtrRef = b; ` `            ``break``; ` `        ``} ` `        ``else` `if` `(b == ``null``) { ` `            ``lastPtrRef = a; ` `            ``break``; ` `        ``} ` `        ``if` `(a.data <= b.data) ` `            ``MoveNode(lastPtrRef, a); ` `        ``else` `            ``MoveNode(lastPtrRef, b); ` ` `  `        ``/* tricky: advance to point to the next ".next" ` `         ``* field */` `        ``lastPtrRef = ((lastPtrRef).next); ` `    ``} ` `    ``return` `(result); ` `} ` ` `  `// This code is contributed by Aditya Kumar (adityakumar129)`

## Python3

 `def` `SortedMerge( a,  b): ` `    ``result ``=` `None``;  ` `         `  `    ``''' point to the last result pointer '''` `    ``lastPtrRef ``=` `result;  ` `         `  `    ``while``(``1``): ` `        ``if` `(a ``=``=` `None``): ` `            ``lastPtrRef ``=` `b;  ` `            ``break``;  ` `        ``elif``(b ``=``=` `None``):  ` `            ``lastPtrRef ``=` `a;  ` `            ``break``;  ` `        ``if``(a.data <``=` `b.data):  ` `            ``MoveNode(lastPtrRef, a);  ` `        ``else``:  ` `            ``MoveNode(lastPtrRef, b);  ` ` `  `        ``''' tricky: advance to point to the next ".next" field '''` `        ``lastPtrRef ``=` `((lastPtrRef).``next``);  ` `    ``return``(result);  ` `   `  ` ``# This code is contributed by umadevi9616  `

## C#

 `Node SortedMerge(Node a, Node b) ` `{ ` `    ``Node result = ``null``; ` ` `  `    ``// Point to the last result pointer ` `    ``Node lastPtrRef = result; ` ` `  `    ``while` `(1) { ` `        ``if` `(a == ``null``) { ` `            ``lastPtrRef = b; ` `            ``break``; ` `        ``} ` `        ``else` `if` `(b == ``null``) { ` `            ``lastPtrRef = a; ` `            ``break``; ` `        ``} ` `        ``if` `(a.data <= b.data) ` `            ``MoveNode(lastPtrRef, a); ` `        ``else` `            ``MoveNode(lastPtrRef, b); ` ` `  `        ``// tricky: advance to point to ` `        ``// the next ".next" field ` `        ``lastPtrRef = ((lastPtrRef).next); ` `    ``} ` `    ``return` `(result); ` `} ` ` `  `// This code is contributed by gauravrajput1`

## Javascript

 ` `

Method 3 (Using Recursion)
Merge is one of those nice recursive problems where the recursive solution code is much cleaner than the iterative code. You probably wouldn’t want to use the recursive version for production code, however, because it will use stack space which is proportional to the length of the lists.

## C++

 `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);  ` `}  ` ` `  `// This code is contributed by rathbhupendra `

## C

 `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); ` `} `

## Java

 `class` `GFG  ` `{ ` `    ``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 : ` `    ``# 10->20->30->40->50 ` `    ``list1 ``=` `LinkedList() ` `    ``list1.append(``10``) ` `    ``list1.append(``20``) ` `    ``list1.append(``30``) ` `    ``list1.append(``40``) ` `    ``list1.append(``50``) ` ` `  `    ``# Create linked list 2 : ` `    ``# 5->15->18->35->60 ` `    ``list2 ``=` `LinkedList() ` `    ``list2.append(``5``) ` `    ``list2.append(``15``) ` `    ``list2.append(``18``) ` `    ``list2.append(``35``) ` `    ``list2.append(``60``) ` ` `  `    ``# 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 : "``, end``=``"") ` `    ``list3.printList()      ` ` `  ` `  `# This code is contributed by 'Shriaknt13'.          `

## C#

 `using` `System; ` ` `  `class` `GFG{ ` ` `  `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; ` `    ``} ` `} ` `} ` ` `  `// This code is contributed by hunter2000`

## Javascript

 ` `

Time Complexity:  Since we are traversing through the two lists fully. So, the time complexity is O(m+n) where m and n are the lengths of the two lists to be merged.

Method 4 (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.

```1) Initialize result list as empty: head = NULL.
2) Let 'a' and 'b' be the heads of first and second list respectively.
3) Reverse both the lists.
4) While (a != NULL and b != NULL)
a) Find the larger of two (Current 'a' and 'b')
b) Insert the larger value of node at the front of result list.
c) Move ahead in the list of larger node.
5) If 'b' becomes NULL before 'a', insert all nodes of 'a'
into result list at the beginning.
6) If 'a' becomes NULL before 'b', insert all nodes of 'b'
into result list at the beginning.  ```

Below is the implementation of 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 ` `#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); ` ` `  `    ``printf``(``"List A before merge: \n"``); ` `    ``printList(a); ` ` `  `    ``printf``(``"\nList B before merge: \n"``); ` `    ``printList(b); ` ` `  `    ``/* merge 2 sorted Linked Lists */` `    ``res = sortedMerge(a, b); ` ` `  `    ``printf``(``"\nMerged Linked List is: \n"``); ` `    ``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 ` `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); ` ` `  `    ``cout << ``"List A before merge: \n"``; ` `    ``printList(a); ` ` `  `    ``cout << ``"\nList B before merge: \n"``; ` `    ``printList(b); ` ` `  `    ``/* merge 2 sorted Linked Lists */` `    ``res = sortedMerge(a, b); ` ` `  `    ``cout << ``"\nMerged 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``); ` ` `  `    ``System.out.print(``"List A before merge: \n"``); ` `    ``printList(a); ` ` `  `    ``System.out.print(``"\nList B before merge: \n"``); ` `    ``printList(b); ` ` `  `    ``/* merge 2 sorted Linked Lists */` `    ``res = sortedMerge(a, b); ` ` `  `    ``System.out.print(``"\nMerged Linked List is: \n"``); ` `    ``printList(res); ` `} ` `} ` ` `  `// This code is contributed by umadevi9616 `

## 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); ` ` `  `    ``Console.Write(``"List A before merge: \n"``); ` `    ``printList(a); ` ` `  `    ``Console.Write(``"\nList B before merge: \n"``); ` `    ``printList(b); ` ` `  `    ``/* merge 2 sorted Linked Lists */` `    ``res = sortedMerge(a, b); ` ` `  `    ``Console.Write(``"\nMerged Linked List is: \n"``); ` `    ``printList(res); ` `} ` `} ` ` `  `// This code is contributed by umadevi9616  `

Output:

```List A before merge:
5 10 15 40
List B before merge:
2 3 20
Merged Linked List is:
2 3 5 10 15 20 40 ```

Time Complexity:  Since we are traversing through the two lists fully. So, the time complexity is O(m+n) where m and n are the lengths of the two lists to be merged.

This method is contributed by Mehul Mathur(mathurmehul01)

This idea is similar to this post.

Please refer below post for simpler implementations :
Merge two sorted lists (in-place)