Merge two unsorted linked lists to get a sorted list - Set 2
Last Updated :
16 Nov, 2023
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;
}
// Java program for the above approach
import java.util.*;
class GFG {
// Linked list node
static class Node {
int data;
Node next;
};
static Node head1, head2;
// Utility function to append key at
// end of linked list
static Node 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;
}
return head;
}
// Utility function to print linkedlist
static void display(Node head)
{
Node temp;
temp = head;
if (temp == null) {
System.out.print("null \n");
}
else {
while (temp != null) {
System.out.print(temp.data);
if (temp.next != null)
System.out.print("->");
temp = temp.next;
}
}
}
// Function to merge two linked lists
static Node MergeLinkedlist()
{
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;
return head1;
}
// Function to merge two unsorted linked
// lists to get a sorted list
static void sortLinkedList()
{
// Function call to merge the two lists
Node head = MergeLinkedlist();
// Declare a vector
Vector<Integer> V = new Vector<>();
Node ptr = head;
// Push all elements into vector
while (ptr != null) {
V.add(ptr.data);
ptr = ptr.next;
}
Collections.sort(V);
// Sort the vector
;
int index = 0;
ptr = head;
// Insert elements in the linked
// list from the vector
while (ptr != null) {
ptr.data = V.get(index);
index++;
ptr = ptr.next;
}
// Display the sorted and
// merged linked list
display(head);
}
// Driver Code
public static void main(String[] args)
{
// Given linked list, L1
head1 = insertNode(head1, 3);
head1 = insertNode(head1, 5);
head1 = insertNode(head1, 1);
// Given linked list, L2
head2 = null;
head2 = insertNode(head2, 6);
head2 = insertNode(head2, 2);
head2 = insertNode(head2, 4);
head2 = insertNode(head2, 9);
// Function Call
sortLinkedList();
}
}
// This code is contributed by umadevi9616
# 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.
// C# program for the above approach
using System;
using System.Collections.Generic;
public class GFG {
// Linked list node
public class Node {
public int data;
public Node next;
};
static Node head1, head2;
// Utility function to append key at
// end of linked list
static Node 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;
}
return head;
}
// Utility function to print linkedlist
static void display(Node head) {
Node temp;
temp = head;
if (temp == null) {
Console.Write("null \n");
} else {
while (temp != null) {
Console.Write(temp.data);
if (temp.next != null)
Console.Write("->");
temp = temp.next;
}
}
}
// Function to merge two linked lists
static Node MergeLinkedlist() {
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;
return head1;
}
// Function to merge two unsorted linked
// lists to get a sorted list
static void sortList()
{
// Function call to merge the two lists
Node head = MergeLinkedlist();
// Declare a vector
List<int> V = new List<int>();
Node ptr = head;
// Push all elements into vector
while (ptr != null) {
V.Add(ptr.data);
ptr = ptr.next;
}
V.Sort();
// Sort the vector
;
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
public static void Main(String[] args)
{
// Given linked list, L1
head1 = insertNode(head1, 3);
head1 = insertNode(head1, 5);
head1 = insertNode(head1, 1);
// Given linked list, L2
head2 = null;
head2 = insertNode(head2, 6);
head2 = insertNode(head2, 2);
head2 = insertNode(head2, 4);
head2 = insertNode(head2, 9);
// Function Call
sortList();
}
}
// This code is contributed by umadevi9616
// 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
Output1->2->3->4->5->6->9
Time Complexity: O((N+M)*log(N+M))
Auxiliary Space: O(N+M)
Another Approach: This code defines a program to merge two unsorted linked lists into a single sorted linked list. The main steps of the program are:
- Define a class called "Node" to represent a node of a linked list. Each node contains an integer value "data" and a pointer "next" to the next node in the list.
- Define a utility function called "insertNode" to insert a new node with a given value at the end of a linked list. This function takes two parameters: a reference to a pointer to the head of the list, and the value to be inserted.
- Define a utility function called "display" to print the elements of a linked list. This function takes a pointer to the head of the list as a parameter and prints the value of each node followed by "->" until the end of the list is reached, and then prints "NULL".
- Define a function called "MergeLinkedlist" to merge two linked lists. This function takes two parameters: a reference to a pointer to the head of the first list, and a pointer to the head of the second list. It first finds the last node of the first list and sets its "next" pointer to the head of the second list.
- Define a function called "sortLinkedList" to merge two unsorted linked lists into a single sorted linked list. This function takes two parameters: pointers to the heads of the two lists. It first calls "MergeLinkedlist" to merge the two lists into a single linked list. Then it creates a vector and pushes all the elements of the merged list into the vector. It sorts the vector using the "sort" function from the <algorithm> library, and then creates a new linked list by inserting the elements of the sorted vector one by one using the "insertNode" function. Finally, it calls "display" to print the sorted linked list.
- In the main function, two unsorted linked lists are created using "insertNode". The "sortLinkedList" function is called with these two lists as parameters.
Below is the implementation of the above approach:
C++
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
// Linked list node
class Node {
public:
int data;
Node* next;
Node(int val) : data(val), next(nullptr) {}
};
// Utility function to append key at
// end of linked list
void insertNode(Node*& head, int x)
{
Node* ptr = new Node(x);
if (head == nullptr) {
head = ptr;
}
else {
Node* temp = head;
while (temp->next != nullptr) {
temp = temp->next;
}
temp->next = ptr;
}
}
// Utility function to print linkedlist
void display(Node* head)
{
if (head == nullptr) {
cout << "NULL" << endl;
}
else {
while (head != nullptr) {
cout << head->data;
if (head->next != nullptr)
cout << "->";
head = head->next;
}
cout << endl;
}
}
// Function to merge two linked lists
void MergeLinkedlist(Node*& head1, Node* head2)
{
Node* ptr = head1;
while (ptr->next != nullptr) {
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* head1, Node* head2)
{
// Function call to merge the two lists
MergeLinkedlist(head1, head2);
// Declare a vector
vector<int> V;
// Push all elements into vector
while (head1 != nullptr) {
V.push_back(head1->data);
head1 = head1->next;
}
// Sort the vector
sort(V.begin(), V.end());
// Insert elements in the linked
// list from the vector
Node* curr = nullptr;
for (int i = 0; i < V.size(); i++) {
if (i == 0) {
curr = new Node(V[i]);
head1 = curr;
}
else {
curr->next = new Node(V[i]);
curr = curr->next;
}
}
// Display the sorted and
// merged linked list
display(head1);
}
// Driver Code
int main()
{
// Given linked list, L1
Node* head1 = nullptr;
insertNode(head1, 3);
insertNode(head1, 5);
insertNode(head1, 1);
// Given linked list, L2
Node* head2 = nullptr;
insertNode(head2, 6);
insertNode(head2, 2);
insertNode(head2, 4);
insertNode(head2, 9);
// Function Call
sortLinkedList(head1, head2);
return 0;
}
// This code is contributed by rudra1807raj
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
class Node {
int data;
Node next;
public Node(int val) {
data = val;
next = null;
}
}
public class MergeAndSortLinkedLists {
// Utility function to append a node at the end of a linked list
static Node insertNode(Node head, int x) {
Node ptr = new Node(x);
if (head == null) {
head = ptr;
} else {
Node temp = head;
while (temp.next != null) {
temp = temp.next;
}
temp.next = ptr;
}
return head;
}
// Utility function to print a linked list
static void display(Node head) {
if (head == null) {
System.out.println("NULL");
} else {
while (head != null) {
System.out.print(head.data);
if (head.next != null) {
System.out.print("->");
}
head = head.next;
}
System.out.println();
}
}
// Function to merge two linked lists
static Node mergeLinkedList(Node head1, Node head2) {
if (head1 == null) {
return head2;
}
Node ptr = head1;
while (ptr.next != null) {
ptr = ptr.next;
}
// Join linked list by placing the address of
// the 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
static void sortLinkedList(Node head1, Node head2) {
// Function call to merge the two lists
head1 = mergeLinkedList(head1, head2);
if (head1 == null) {
System.out.println("Empty List");
return;
}
// Declare a list
List<Integer> list = new ArrayList<>();
// Push all elements into the list
Node current = head1;
while (current != null) {
list.add(current.data);
current = current.next;
}
// Sort the list
Collections.sort(list);
// Insert elements in the linked list from the list
current = head1;
for (int i = 0; i < list.size(); i++) {
current.data = list.get(i);
current = current.next;
}
// Display the sorted and merged linked list
display(head1);
}
// Driver Code
public static void main(String[] args) {
// Given linked list, L1
Node head1 = null;
head1 = insertNode(head1, 3);
head1 = insertNode(head1, 5);
head1 = insertNode(head1, 1);
// Given linked list, L2
Node head2 = null;
head2 = insertNode(head2, 6);
head2 = insertNode(head2, 2);
head2 = insertNode(head2, 4);
head2 = insertNode(head2, 9);
// Function Call
sortLinkedList(head1, head2);
}
}
# Linked list node
class Node:
def __init__(self, val):
self.data = val
self.next = None
# Utility function to append key at
# end of linked list
def insertNode(head, x):
ptr = Node(x)
if head is None:
head = ptr
else:
temp = head
while temp.next is not None:
temp = temp.next
temp.next = ptr
return head
# Utility function to print linkedlist
def display(head):
if head is None:
print("NULL")
else:
while head is not None:
print(head.data, end="")
if head.next is not None:
print("->", end="")
head = head.next
print()
# Function to merge two linked lists
def MergeLinkedlist(head1, head2):
ptr = head1
while ptr.next is not None:
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
def sortLinkedList(head1, head2):
# Function call to merge the two lists
MergeLinkedlist(head1, head2)
# Declare a vector
V = []
# Push all elements into vector
while head1 is not None:
V.append(head1.data)
head1 = head1.next
# Sort the vector
V.sort()
# Insert elements in the linked
# list from the vector
curr = None
for i in range(len(V)):
if i == 0:
curr = Node(V[i])
head1 = curr
else:
curr.next = Node(V[i])
curr = curr.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 shiv1o43g
using System;
using System.Collections.Generic;
using System.Linq;
// Linked list node
class Node {
public int data;
public Node next;
public Node(int val) {
data = val;
next = null;
}
}
class Program {
// Utility function to append key at
// end of linked list
static void insertNode(ref Node head, int x) {
Node ptr = new Node(x);
if (head == null) {
head = ptr;
}
else {
Node temp = head;
while (temp.next != null) {
temp = temp.next;
}
temp.next = ptr;
}
}
// Utility function to print linkedlist
static void display(Node head) {
if (head == null) {
Console.WriteLine("NULL");
}
else {
while (head != null) {
Console.Write(head.data);
if (head.next != null)
Console.Write("->");
head = head.next;
}
Console.WriteLine();
}
}
// Function to merge two linked lists
static void MergeLinkedlist(ref Node head1, Node head2) {
Node 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
static void sortLinkedList(Node head1, Node head2) {
// Function call to merge the two lists
MergeLinkedlist(ref head1, head2);
// Declare a list
List<int> lst = new List<int>();
// Push all elements into the list
while (head1 != null) {
lst.Add(head1.data);
head1 = head1.next;
}
// Sort the list
lst.Sort();
// Insert elements in the linked
// list from the list
Node curr = null;
for (int i = 0; i < lst.Count; i++) {
if (i == 0) {
curr = new Node(lst[i]);
head1 = curr;
}
else {
curr.next = new Node(lst[i]);
curr = curr.next;
}
}
// Display the sorted and
// merged linked list
display(head1);
}
// Driver Code
static void Main(string[] args) {
// Given linked list, L1
Node head1 = null;
insertNode(ref head1, 3);
insertNode(ref head1, 5);
insertNode(ref head1, 1);
// Given linked list, L2
Node head2 = null;
insertNode(ref head2, 6);
insertNode(ref head2, 2);
insertNode(ref head2, 4);
insertNode(ref head2, 9);
// Function Call
sortLinkedList(head1, head2);
Console.ReadLine();
}
}
//This code is contributed by rudra1807raj
class Node {
constructor(val) {
this.data = val;
this.next = null;
}
}
// Utility function to append a node at the end of a linked list
function insertNode(head, x) {
const ptr = new Node(x);
if (head === null) {
head = ptr;
} else {
let temp = head;
while (temp.next !== null) {
temp = temp.next;
}
temp.next = ptr;
}
return head;
}
// Utility function to print a linked list
function display(head) {
if (head === null) {
console.log("NULL");
} else {
while (head !== null) {
process.stdout.write(head.data.toString());
if (head.next !== null) {
process.stdout.write("->");
}
head = head.next;
}
console.log();
}
}
// Function to merge two linked lists
function mergeLinkedList(head1, head2) {
if (head1 === null) {
return head2;
}
let ptr = head1;
while (ptr.next !== null) {
ptr = ptr.next;
}
// Join linked list by placing the address of
// the 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);
if (head1 === null) {
console.log("Empty List");
return;
}
// Declare an array
const list = [];
// Push all elements into the array
let current = head1;
while (current !== null) {
list.push(current.data);
current = current.next;
}
// Sort the array
list.sort();
// Insert elements in the linked list from the array
current = head1;
for (let i = 0; i < list.length; i++) {
current.data = list[i];
current = current.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);
Output1->2->3->4->5->6->9
Time Complexity: The time complexity of the given algorithm is O(n log n) where n is the total number of elements in the two linked lists. This is because the algorithm involves sorting the elements using the sort function, which has a time complexity of O(n log n). In addition, iterating over the linked lists to populate the vector and then the sorted linked list takes O(n) time.
Auxiliary Space: The space complexity of the given algorithm is O(n) where n is the total number of elements in the two linked lists. This is because we are creating a vector of size n to store all the elements of the linked list before sorting. In addition, we are creating a new linked list of size n to store the sorted elements. However, we are not using any extra space proportional to the size of the input beyond the two linked lists and the vector used for sorting, so the space complexity is O(n).
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