# Merge two sorted linked lists

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’.

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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 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 tail. When
we are done, the result is in dummy.next.

Below image is a dry run of the above approach: Below is the implemenation 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}  ` ` `  `Affter 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 beginging of the linked list */` `void` `push(Node** head_ref, ``int` `new_data)  ` `{  ` `    ``/* allocate node */` `    ``Node* new_node = ``new` `Node(); ` ` `  `    ``/* put in the data */` `    ``new_node->data = new_data;  ` ` `  `    ``/* link the old list off the new node */` `    ``new_node->next = (*head_ref);  ` ` `  `    ``/* move the head to point to the new node */` `    ``(*head_ref) = new_node;  ` `}  ` ` `  `/* 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} ` ` `  `   ``Affter 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 beginging of the ` `   ``linked list */` `void` `push(``struct` `Node** head_ref, ``int` `new_data) ` `{ ` `    ``/* allocate node */` `    ``struct` `Node* new_node = ` `        ``(``struct` `Node*) ``malloc``(``sizeof``(``struct` `Node)); ` ` `  `    ``/* put in the data  */` `    ``new_node->data  = new_data; ` ` `  `    ``/* link the old list off the new node */` `    ``new_node->next = (*head_ref); ` ` `  `    ``/* move the head to point to the new node */` `    ``(*head_ref)    = new_node; ` `} ` ` `  `/* Function to print nodes in a given linked list */` `void` `printList(``struct` `Node *node) ` `{ ` `    ``while` `(node!=NULL) ` `    ``{ ` `        ``printf``(``"%d "``, node->data); ` `        ``node = node->next; ` `    ``} ` `} ` ` `  `/* Drier 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 `

## 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 `

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++

 `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 rathbhupendra `

## 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); ` `} `

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'.          `

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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