Merge two sorted linked lists

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.
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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: 
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/* C++ program to merge two sorted linked lists */#include <bits/stdc++.h>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 increasingorder, and splices their nodes togetherto 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 thenode 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<<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 program to merge two sorted linked lists */#include<stdio.h>#include<stdlib.h>#include<assert.h>  /* 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 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 Codepublic 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
""" Python program to merge twosorted linked lists """    # Linked List Nodeclass Node:    def __init__(self, data):        self.data = data        self.next = None    # Create & Handle List operationsclass 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 listdef 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 listslistA = LinkedList()listB = LinkedList()  # Add elements to the list in sorted orderlistA.addToList(5)listA.addToList(10)listA.addToList(15)  listB.addToList(2)listB.addToList(3)listB.addToList(20)  # Call the merge functionlistA.head = mergeLists(listA.head, listB.head)  # Display merged listprint("Merged Linked List is:")listA.printList()  """ This code is contributedby Debidutta Rath """
/* 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
<script>  class Node{    constructor(d)    {        this.data=d;        this.next = null;    }}  class LinkedList{  constructor(){    this.head=null;}    addToTheLast(node){    if (this.head == null)    {        this.head = node;    }    else    {        let temp = this.head;        while (temp.next != null)            temp = temp.next;        temp.next = node;    }}  printList(){    let temp = this.head;    while (temp != null)    {        document.write(temp.data + " ");        temp = temp.next;    }    document.write("<br>");}}  function sortedMerge(headA,headB){    /* a dummy first node to       hang the result on */    let dummyNode = new Node(0);           /* tail points to the    last result node */    let 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;}    /* Let us create two sorted linked       lists to test the methods       Created lists:           llist1: 5->10->15,           llist2: 2->3->20    */let llist1 = new LinkedList();let llist2 = new LinkedList();  // 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 = sortedMerge(llist1.head,                                    llist2.head);document.write("Merged Linked List is:<br>")llist1.printList();      // This code is contributed by patel2127</script>
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). 
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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)
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)
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)
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 
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
<script>  function SortedMerge(a,b){    let result = null;       /* point to the last result pointer */let 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 rag2127  </script>
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.  
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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
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);}
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 program merge two sorted linked# in third linked list using recursive.  # Node classclass Node:    def __init__(self, data):        self.data = data        self.next = None    # Constructor to initialize the node objectclass 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 Functionif __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'.         
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
<script>      function SortedMerge( A,  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 contributed by umadevi9616 </script>
 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.
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Java
C#
/*Given two sorted linked lists consisting of N and M nodesrespectively. The task is to merge both of the list(in-place) and return head of the merged list.*/#include <stdio.h>#include <stdlib.h>  /* Link list Node */typedef struct Node {    int key;    struct Node* next;}Node;  // Function to reverse a given Linked List using RecursionNode* 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)
/*Given two sorted linked lists consisting of N and M nodesrespectively. The task is to merge both of the list(in-place) and return head of the merged list.*/#include <iostream>using namespace std;  /* Link list Node */struct Node {    int key;    struct Node* next;};  // Function to reverse a given Linked List using RecursionNode* 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)
/*Given two sorted linked lists consisting of N and M nodesrespectively. 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 Recursionstatic 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
/*Given two sorted linked lists consisting of N and M nodesrespectively. 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 Recursionstatic 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)Source: http://cslibrary.stanford.edu/105/LinkedListProblems.pdfPlease write comments if you find the above code/algorithm incorrect, or find better ways to solve the same problem.
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Zoho
Flipkart
Accolite
Amazon
Microsoft
Samsung
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Oracle
Brocade
Synopsys
OATS Systems
Belzabar
Linked List
Merge Sort
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