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Merge two unsorted linked lists to get a sorted list – Set 2

  • Difficulty Level : Easy
  • Last Updated : 19 Jul, 2021

Given two unsorted Linked Lists, L1 of N nodes and L2 of M nodes, the task is to merge them to get a sorted singly linked list.

Examples:

Input: L1 = 3→5→1, L2 = 6→2→4→9
Output: 1→2→3→4→5→6→9

Input: L1 = 1→5→2, L2 = 3→7
Output: 1→2→3→5→7

Note: A Memory Efficient approach that solves this problem in O(1) Auxiliary Space has been approached here,



Approach: The idea is to use an auxiliary array to store the elements of both the linked lists and then sort the array in increasing order. And finally, insert all the elements back into the linked list. Follow the steps below to solve the problem:

Below is an implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Linked list node
class Node {
public:
    int data;
    Node* next;
};
 
// Utility function to append key at
// end of linked list
void insertNode(Node** head, int x)
{
    Node* ptr = new Node;
    ptr->data = x;
    ptr->next = NULL;
    if (*head == NULL) {
        *head = ptr;
    }
    else {
        Node* temp;
        temp = *head;
        while (temp->next != NULL) {
            temp = temp->next;
        }
        temp->next = ptr;
    }
}
 
// Utility function to print linkedlist
void display(Node** head)
{
    Node* temp;
    temp = *head;
    if (temp == NULL) {
        cout << "NULL \n";
    }
    else {
        while (temp != NULL) {
            cout << temp->data;
            if (temp->next != NULL)
                cout << "->";
            temp = temp->next;
        }
    }
}
 
// Function to merge two linked lists
void MergeLinkedlist(Node** head1, Node** head2)
{
    Node* ptr;
    ptr = *head1;
    while (ptr->next != NULL) {
        ptr = ptr->next;
    }
 
    // Join linked list by placing address of
    // first node of L2 in the last node of L1
    ptr->next = *head2;
}
 
// Function to merge two unsorted linked
// lists to get a sorted list
void sortLinkedList(Node** head, Node** head1)
{
    // Function call to merge the two lists
    MergeLinkedlist(head, head1);
 
    // Declare a vector
    vector<int> V;
    Node* ptr = *head;
 
    // Push all elements into vector
    while (ptr != NULL) {
        V.push_back(ptr->data);
        ptr = ptr->next;
    }
 
    // Sort the vector
    sort(V.begin(), V.end());
    int index = 0;
    ptr = *head;
 
    // Insert elements in the linked
    // list from the vector
    while (ptr != NULL) {
        ptr->data = V[index];
        index++;
        ptr = ptr->next;
    }
 
    // Display the sorted and
    // merged linked list
    display(head);
}
 
// Driver Code
int main()
{
    // Given linked list, L1
    Node* head1 = NULL;
    insertNode(&head1, 3);
    insertNode(&head1, 5);
    insertNode(&head1, 1);
 
    // Given linked list, L2
    Node* head2 = NULL;
    insertNode(&head2, 6);
    insertNode(&head2, 2);
    insertNode(&head2, 4);
    insertNode(&head2, 9);
 
    // Function Call
    sortLinkedList(&head1, &head2);
 
    return 0;
}

Python3




# Py program for the above approach
 
# Linked list node
class Node:
    def __init__(self, d):
        self.data = d
        self.next = None
 
# Utility function to append key at
# end of linked list
def insertNode(head, x):
    ptr = Node(x)
    if (head == None):
        head = ptr
    else:
        temp = head
        while (temp.next != None):
            temp = temp.next
        temp.next = ptr
    return head
 
# Utility function to print linkedlist
def display(head):
    temp = head
    if (temp == None):
        print ("None")
    else:
        while (temp.next != None):
            print (temp.data,end="->")
            if (temp.next != None):
                print ("",end="")
            temp = temp.next
        print(temp.data)
 
# Function to merge two linked lists
def MergeLinkedlist(head1, head2):
    ptr = head1
    while (ptr.next != None):
        ptr = ptr.next
 
    # Join linked list by placing address of
    # first node of L2 in the last node of L1
    ptr.next = head2
 
    return head1
 
# Function to merge two unsorted linked
# lists to get a sorted list
def sortLinkedList(head1, head2):
    # Function call to merge the two lists
    head1 = MergeLinkedlist(head1, head2)
 
    # Declare a vector
    V = []
    ptr = head1
 
    # Push all elements into vector
    while (ptr != None):
        V.append(ptr.data)
        ptr = ptr.next
 
    # Sort the vector
    V = sorted(V)
    index = 0
    ptr = head1
 
    # Insert elements in the linked
    # list from the vector
    while (ptr != None):
        ptr.data = V[index]
        index += 1
        ptr = ptr.next
 
    # Display the sorted and
    # merged linked list
    display(head1)
 
# Driver Code
if __name__ == '__main__':
    # Given linked list, L1
    head1 = None
    head1 = insertNode(head1, 3)
    head1 = insertNode(head1, 5)
    head1 = insertNode(head1, 1)
 
 
    # Given linked list, L2
    head2 = None
    head2 = insertNode(head2, 6)
    head2 = insertNode(head2, 2)
    head2 = insertNode(head2, 4)
    head2 = insertNode(head2, 9)
 
    # Function Call
    sortLinkedList(head1, head2)
 
# This code is contributed by mohit kumar 29.

Javascript




// Javascript program for the above approach
 
// Linked list node
class Node {
    constructor(d) {
        this.data = d
        this.next = null
    }
}
 
// Utility function to append key at
// end of linked list
function insertNode(head, x) {
    ptr = new Node(x)
    if (head == null) {
        head = ptr
    } else {
        temp = head
        while (temp.next != null) {
            temp = temp.next
        }
        temp.next = ptr
    }
    return head
}
 
// Utility function to print linkedlist
function display(head) {
    let temp = head
    if (temp == null) {
        document.write("null")
    }
    else {
        while (temp.next != null) {
            document.write(temp.data, end = "->")
            if (temp.next != null)
                document.write("", end = "")
            temp = temp.next
        }
        document.write(temp.data)
    }
}
 
// Function to merge two linked lists
function MergeLinkedlist(head1, head2) {
    let ptr = head1
    while (ptr.next != null)
        ptr = ptr.next
 
    // Join linked list by placing address of
    // first node of L2 in the last node of L1
    ptr.next = head2
 
    return head1
}
 
// Function to merge two unsorted linked
// lists to get a sorted list
function sortLinkedList(head1, head2) {
    // Function call to merge the two lists
    head1 = MergeLinkedlist(head1, head2)
 
    // Declare a vector
    let V = []
    let ptr = head1
 
    // Push all elements into vector
    while (ptr != null) {
        V.push(ptr.data)
        ptr = ptr.next
    }
 
    // Sort the vector
    V = V.sort((a, b) => a - b)
    index = 0
    ptr = head1
 
    // Insert elements in the linked
    // list from the vector
    while (ptr != null) {
        ptr.data = V[index]
        index += 1
        ptr = ptr.next
    }
 
    // Display the sorted and
    // merged linked list
    display(head1)
}
 
// Driver Code
 
// Given linked list, L1
let head1 = null;
head1 = insertNode(head1, 3)
head1 = insertNode(head1, 5)
head1 = insertNode(head1, 1)
 
 
// Given linked list, L2
let head2 = null
head2 = insertNode(head2, 6)
head2 = insertNode(head2, 2)
head2 = insertNode(head2, 4)
head2 = insertNode(head2, 9)
 
// Function Call
sortLinkedList(head1, head2)
 
// This code is contributed by gfgking
Output
1->2->3->4->5->6->9

Time Complexity: O((N+M)*log(N+M))
Auxiliary Space: O(N+M)

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