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Merge two sorted linked lists

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  • Difficulty Level : Medium
  • Last Updated : 28 Oct, 2022
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Given two sorted linked lists consisting of N and M nodes respectively. The task is to merge both of the lists (in place) and return the head of the merged list.

Examples:

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

Input: a: 1->1, b: 2->4
Output: 1->1->2->4

Complete Interview Preparation - GFG

Merge two sorted linked lists using Dummy Nodes:

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

Follow the below illustration for a better understanding:

Illustration:

Merge-Two-Sorted-LinkedLists1

Follow the steps below to solve the problem:

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

Below is the implementation of the above approach: 

C++




/* C++ program to merge two sorted linked lists */
#include <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 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 << node->data << " ";
        node = node->next;
    }
}
 
/* Driver code*/
int main()
{
    /* Start with the empty list */
    Node* res = NULL;
    Node* a = NULL;
    Node* b = NULL;
 
    /* Let us create two sorted linked lists
    to test the functions
    Created lists, a: 5->10->15, b: 2->3->20 */
    push(&a, 15);
    push(&a, 10);
    push(&a, 5);
 
    push(&b, 20);
    push(&b, 3);
    push(&b, 2);
 
    /* Remove duplicates from linked list */
    res = SortedMerge(a, b);
 
    cout << "Merged Linked List is: \n";
    printList(res);
 
    return 0;
}
 
// This code is contributed by rathbhupendra


C




/* C program to merge two sorted linked lists */
#include <assert.h>
#include <stdio.h>
#include <stdlib.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




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


Python3




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


C#




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


Javascript




<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 

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

Merge two sorted linked lists using Recursion:

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

Follow the steps below to solve the problem:

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

Below is the implementation of the above approach.

C++




/* C++ program to merge two sorted linked lists */
#include <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 increasing
order, and splices their nodes together
to make one big sorted list which
is returned. */
Node* SortedMerge(Node* a, Node* b)
{
    Node* result = NULL;
 
    /* Base cases */
    if (a == NULL)
        return (b);
    else if (b == NULL)
        return (a);
 
    /* Pick either a or b, and recur */
    if (a->data <= b->data) {
        result = a;
        result->next = SortedMerge(a->next, b);
    }
    else {
        result = b;
        result->next = SortedMerge(a, b->next);
    }
    return (result);
}
 
/* 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 << node->data << " ";
        node = node->next;
    }
}
 
/* Driver code*/
int main()
{
    /* Start with the empty list */
    Node* res = NULL;
    Node* a = NULL;
    Node* b = NULL;
 
    /* Let us create two sorted linked lists
    to test the functions
    Created lists, a: 5->10->15, b: 2->3->20 */
    push(&a, 15);
    push(&a, 10);
    push(&a, 5);
 
    push(&b, 20);
    push(&b, 3);
    push(&b, 2);
 
    /* Remove duplicates from linked list */
    res = SortedMerge(a, b);
 
    cout << "Merged Linked List is: \n";
    printList(res);
 
    return 0;
}
 
// This code is contributed by rathbhupendra


C




/* C program to merge two sorted linked lists */
#include <assert.h>
#include <stdio.h>
#include <stdlib.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)
{
    struct Node* result = NULL;
 
    /* Base cases */
    if (a == NULL)
        return (b);
    else if (b == NULL)
        return (a);
 
    /* Pick either a or b, and recur */
    if (a->data <= b->data) {
        result = a;
        result->next = SortedMerge(a->next, b);
    }
    else {
        result = b;
        result->next = SortedMerge(a, b->next);
    }
    return (result);
}
 
/* UTILITY FUNCTIONS */
/* MoveNode() function takes the node from the front of the
   source, and move it to the front of the dest.
   It is an error to call this with the source list empty.
 
   Before calling MoveNode():
   source == {1, 2, 3}
   dest == {1, 2, 3}
 
   After calling MoveNode():
   source == {2, 3}
   dest == {1, 1, 2, 3} */
void MoveNode(struct Node** destRef,
              struct Node** sourceRef)
{
    /* the front source node  */
    struct Node* newNode = *sourceRef;
    assert(newNode != NULL);
 
    /* Advance the source pointer */
    *sourceRef = newNode->next;
 
    /* Link the old dest off the new node */
    newNode->next = *destRef;
 
    /* Move dest to point to the new node */
    *destRef = newNode;
}
 
/* Function to insert a node at the beginning of the
   linked list */
void push(struct Node** head_ref, int new_data)
{
    /* allocate node */
    struct Node* new_node
        = (struct Node*)malloc(sizeof(struct Node));
 
    /* put in the data  */
    new_node->data = new_data;
 
    /* link the old list 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()
    {
        System.out.println("Merged Linked List is:");
        Node temp = head;
        while (temp != null) {
            System.out.print(temp.data + " ");
            temp = temp.next;
        }
        System.out.println();
    }
 
    // Driver Code
    public static void main(String args[])
    {
        /* Let us create two sorted linked
           lists to test the methods
           Created lists:
               llist1: 5->10->15,
               llist2: 2->3->20
        */
        MergeLists llist1 = new MergeLists();
        MergeLists llist2 = new MergeLists();
 
        // Node head1 = new Node(5);
        llist1.addToTheLast(new Node(5));
        llist1.addToTheLast(new Node(10));
        llist1.addToTheLast(new Node(15));
 
        // Node head2 = new Node(2);
        llist2.addToTheLast(new Node(2));
        llist2.addToTheLast(new Node(3));
        llist2.addToTheLast(new Node(20));
 
        llist1.head = new Gfg().sortedMerge(llist1.head,
                                            llist2.head);
        llist1.printList();
    }
}
 
class Gfg {
    /* Takes two lists sorted in
    increasing order, and splices
    their nodes together to make
    one big sorted list which is
    returned. */
    public Node sortedMerge(Node A, Node B)
    {
 
        if (A == null)
            return B;
        if (B == null)
            return A;
 
        if (A.data < B.data) {
            A.next = sortedMerge(A.next, B);
            return A;
        }
        else {
            B.next = sortedMerge(A, B.next);
            return B;
        }
    }
}
 
// This code is contributed by Tuhin Das


Python3




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


C#




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


Javascript




<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>


Output

Merged Linked List is: 
2 3 5 10 15 20 

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

Merge two sorted linked lists by Reversing the Lists:

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

Follow the steps below to solve the problem:

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

Below is the implementation of the above solution.

C++




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


C




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


Java




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


C#




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


Output

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

Time Complexity: O(M+N) where M and N are the lengths of the two lists to be merged. 
Auxiliary Space: O(1)

This method is contributed by Mehul Mathur(mathurmehul01)

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


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