Merge two unsorted linked lists to get a sorted list
Last Updated :
15 Jul, 2025
Given two unsorted Linked List, the task is to merge them to get a sorted singly linked list.
Examples:
Input: List 1 = 3 -> 1 -> 5, List 2 = 6-> 2 -> 4
Output: 1 -> 2 -> 3 -> 4 -> 5 -> 6
Input: List 1 = 4 -> 7 -> 5, List 2 = 2-> 1 -> 8 -> 1
Output: 1 -> 1 -> 2 -> 4 -> 5 -> 7 -> 8
Naive Approach: The naive approach is to sort the given linked lists and then merge the two sorted linked lists together into one list in increasing order.
To solve the problem mentioned above the naive method is to sort the two linked lists individually and merge the two linked lists together into one list which is in increasing order.
Efficient Approach: To optimize the above method we will concatenate the two linked lists and then sort it using any sorting algorithm. Below are the steps:
- Concatenate the two lists by traversing the first list until we reach it's a tail node and then point the next of the tail node to the head node of the second list. Store this concatenated list in the first list.
- Sort the above-merged linked list. Here, we will use a bubble sort. So, if node->next->data is less then node->data, then swap the data of the two adjacent nodes.
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Create structure for a node
struct node {
int data;
node* next;
};
// Function to print the linked list
void setData(node* head)
{
node* tmp;
// Store the head of the linked
// list into a temporary node*
// and iterate
tmp = head;
while (tmp != NULL) {
cout << tmp->data
<< " -> ";
tmp = tmp->next;
}
}
// Function takes the head of the
// LinkedList and the data as
// argument and if no LinkedList
// exists, it creates one with the
// head pointing to first node.
// If it exists already, it appends
// given node at end of the last node
node* getData(node* head, int num)
{
// Create a new node
node* temp = new node;
node* tail = head;
// Insert data into the temporary
// node and point it's next to NULL
temp->data = num;
temp->next = NULL;
// Check if head is null, create a
// linked list with temp as head
// and tail of the list
if (head == NULL) {
head = temp;
tail = temp;
}
// Else insert the temporary node
// after the tail of the existing
// node and make the temporary node
// as the tail of the linked list
else {
while (tail != NULL) {
if (tail->next == NULL) {
tail->next = temp;
tail = tail->next;
}
tail = tail->next;
}
}
// Return the list
return head;
}
// Function to concatenate the two lists
node* mergelists(node** head1,
node** head2)
{
node* tail = *head1;
// Iterate through the head1 to find the
// last node join the next of last node
// of head1 to the 1st node of head2
while (tail != NULL) {
if (tail->next == NULL
&& head2 != NULL) {
tail->next = *head2;
break;
}
tail = tail->next;
}
// return the concatenated lists as a
// single list - head1
return *head1;
}
// Sort the linked list using bubble sort
void sortlist(node** head1)
{
node* curr = *head1;
node* temp = *head1;
// Compares two adjacent elements
// and swaps if the first element
// is greater than the other one.
while (curr->next != NULL) {
temp = curr->next;
while (temp != NULL) {
if (temp->data < curr->data) {
int t = temp->data;
temp->data = curr->data;
curr->data = t;
}
temp = temp->next;
}
curr = curr->next;
}
}
// Driver Code
int main()
{
node* head1 = new node;
node* head2 = new node;
head1 = NULL;
head2 = NULL;
// Given Linked List 1
head1 = getData(head1, 4);
head1 = getData(head1, 7);
head1 = getData(head1, 5);
// Given Linked List 2
head2 = getData(head2, 2);
head2 = getData(head2, 1);
head2 = getData(head2, 8);
head2 = getData(head2, 1);
// Merge the two lists
// in a single list
head1 = mergelists(&head1,
&head2);
// Sort the unsorted merged list
sortlist(&head1);
// Print the final
// sorted merged list
setData(head1);
return 0;
}
// Java program for
// the above approach
class GFG{
static node head1 = null;
static node head2 = null;
// Create structure for a node
static class node
{
int data;
node next;
};
// Function to print
// the linked list
static void setData(node head)
{
node tmp;
// Store the head of the linked
// list into a temporary node
// and iterate
tmp = head;
while (tmp != null)
{
System.out.print(tmp.data + " -> ");
tmp = tmp.next;
}
}
// Function takes the head of the
// LinkedList and the data as
// argument and if no LinkedList
// exists, it creates one with the
// head pointing to first node.
// If it exists already, it appends
// given node at end of the last node
static node getData(node head, int num)
{
// Create a new node
node temp = new node();
node tail = head;
// Insert data into the temporary
// node and point it's next to null
temp.data = num;
temp.next = null;
// Check if head is null, create a
// linked list with temp as head
// and tail of the list
if (head == null)
{
head = temp;
tail = temp;
}
// Else insert the temporary node
// after the tail of the existing
// node and make the temporary node
// as the tail of the linked list
else
{
while (tail != null)
{
if (tail.next == null)
{
tail.next = temp;
tail = tail.next;
}
tail = tail.next;
}
}
// Return the list
return head;
}
// Function to concatenate
// the two lists
static node mergelists()
{
node tail = head1;
// Iterate through the
// head1 to find the
// last node join the
// next of last node
// of head1 to the
// 1st node of head2
while (tail != null)
{
if (tail.next == null &&
head2 != null)
{
tail.next = head2;
break;
}
tail = tail.next;
}
// return the concatenated
// lists as a single list - head1
return head1;
}
// Sort the linked list
// using bubble sort
static void sortlist()
{
node curr = head1;
node temp = head1;
// Compares two adjacent elements
// and swaps if the first element
// is greater than the other one.
while (curr.next != null)
{
temp = curr.next;
while (temp != null)
{
if (temp.data < curr.data)
{
int t = temp.data;
temp.data = curr.data;
curr.data = t;
}
temp = temp.next;
}
curr = curr.next;
}
}
// Driver Code
public static void main(String[] args)
{
// Given Linked List 1
head1 = getData(head1, 4);
head1 = getData(head1, 7);
head1 = getData(head1, 5);
// Given Linked List 2
head2 = getData(head2, 2);
head2 = getData(head2, 1);
head2 = getData(head2, 8);
head2 = getData(head2, 1);
// Merge the two lists
// in a single list
head1 = mergelists();
// Sort the unsorted merged list
sortlist();
// Print the final
// sorted merged list
setData(head1);
}
}
// This code is contributed by shikhasingrajput
# Python3 program for the
# above approach
# Create structure for a node
class node:
def __init__(self, x):
self.data = x
self.next = None
# Function to print the linked
# list
def setData(head):
# Store the head of the
# linked list into a
# temporary node* and
# iterate
tmp = head
while (tmp != None):
print(tmp.data,
end = " -> ")
tmp = tmp.next
# Function takes the head of the
# LinkedList and the data as
# argument and if no LinkedList
# exists, it creates one with the
# head pointing to first node.
# If it exists already, it appends
# given node at end of the last node
def getData(head, num):
# Create a new node
temp = node(-1)
tail = head
# Insert data into the temporary
# node and point it's next to NULL
temp.data = num
temp.next = None
# Check if head is null, create a
# linked list with temp as head
# and tail of the list
if (head == None):
head = temp
tail = temp
# Else insert the temporary node
# after the tail of the existing
# node and make the temporary node
# as the tail of the linked list
else:
while (tail != None):
if (tail.next == None):
tail.next = temp
tail = tail.next
tail = tail.next
# Return the list
return head
# Function to concatenate the
# two lists
def mergelists(head1,head2):
tail = head1
# Iterate through the head1 to
# find the last node join the
# next of last node of head1
# to the 1st node of head2
while (tail != None):
if (tail.next == None
and head2 != None):
tail.next =head2
break
tail = tail.next
# return the concatenated
# lists as a single list
# - head1
return head1
# Sort the linked list using
# bubble sort
def sortlist(head1):
curr = head1
temp = head1
# Compares two adjacent elements
# and swaps if the first element
# is greater than the other one.
while (curr.next != None):
temp = curr.next
while (temp != None):
if (temp.data < curr.data):
t = temp.data
temp.data = curr.data
curr.data = t
temp = temp.next
curr = curr.next
# Driver Code
if __name__ == '__main__':
head1 = node(-1)
head2 = node(-1)
head1 = None
head2 = None
# Given Linked List 1
head1 = getData(head1, 4)
head1 = getData(head1, 7)
head1 = getData(head1, 5)
# Given Linked List 2
head2 = getData(head2, 2)
head2 = getData(head2, 1)
head2 = getData(head2, 8)
head2 = getData(head2, 1)
# Merge the two lists
# in a single list
head1 = mergelists(head1,head2)
# Sort the unsorted merged list
sortlist(head1)
# Print the final
# sorted merged list
setData(head1)
# This code is contributed by Mohit Kumar 29
// C# program for
// the above approach
using System;
class GFG{
static node head1 = null;
static node head2 = null;
// Create structure for a node
class node
{
public int data;
public node next;
};
// Function to print
// the linked list
static void setData(node head)
{
node tmp;
// Store the head of the linked
// list into a temporary node
// and iterate
tmp = head;
while (tmp != null)
{
Console.Write(tmp.data + " -> ");
tmp = tmp.next;
}
}
// Function takes the head of
//the List and the data as
// argument and if no List
// exists, it creates one with the
// head pointing to first node.
// If it exists already, it appends
// given node at end of the last node
static node getData(node head, int num)
{
// Create a new node
node temp = new node();
node tail = head;
// Insert data into the temporary
// node and point it's next to null
temp.data = num;
temp.next = null;
// Check if head is null, create a
// linked list with temp as head
// and tail of the list
if (head == null)
{
head = temp;
tail = temp;
}
// Else insert the temporary node
// after the tail of the existing
// node and make the temporary node
// as the tail of the linked list
else
{
while (tail != null)
{
if (tail.next == null)
{
tail.next = temp;
tail = tail.next;
}
tail = tail.next;
}
}
// Return the list
return head;
}
// Function to concatenate
// the two lists
static node mergelists()
{
node tail = head1;
// Iterate through the
// head1 to find the
// last node join the
// next of last node
// of head1 to the
// 1st node of head2
while (tail != null)
{
if (tail.next == null &&
head2 != null)
{
tail.next = head2;
break;
}
tail = tail.next;
}
// return the concatenated
// lists as a single list - head1
return head1;
}
// Sort the linked list
// using bubble sort
static void sortlist()
{
node curr = head1;
node temp = head1;
// Compares two adjacent elements
// and swaps if the first element
// is greater than the other one.
while (curr.next != null)
{
temp = curr.next;
while (temp != null)
{
if (temp.data < curr.data)
{
int t = temp.data;
temp.data = curr.data;
curr.data = t;
}
temp = temp.next;
}
curr = curr.next;
}
}
// Driver Code
public static void Main(String[] args)
{
// Given Linked List 1
head1 = getData(head1, 4);
head1 = getData(head1, 7);
head1 = getData(head1, 5);
// Given Linked List 2
head2 = getData(head2, 2);
head2 = getData(head2, 1);
head2 = getData(head2, 8);
head2 = getData(head2, 1);
// Merge the two lists
// in a single list
head1 = mergelists();
// Sort the unsorted merged list
sortlist();
// Print the final
// sorted merged list
setData(head1);
}
}
// This code is contributed by Amit Katiyar
<script>
// javascript program for
// the above approach
var head1 = null;
var head2 = null;
// Create structure for a node
class node {
constructor(){
this.data=0;
this.next = null;
}
}
// Function to print
// the linked list
function setData( head) {
var tmp;
// Store the head of the linked
// list into a temporary node
// and iterate
tmp = head;
while (tmp != null) {
document.write(tmp.data + " -> ");
tmp = tmp.next;
}
}
// Function takes the head of the
// LinkedList and the data as
// argument and if no LinkedList
// exists, it creates one with the
// head pointing to first node.
// If it exists already, it appends
// given node at end of the last node
function getData( head , num) {
// Create a new node
temp = new node();
var tail = head;
// Insert data into the temporary
// node and point it's next to null
temp.data = num;
temp.next = null;
// Check if head is null, create a
// linked list with temp as head
// and tail of the list
if (head == null) {
head = temp;
tail = temp;
}
// Else insert the temporary node
// after the tail of the existing
// node and make the temporary node
// as the tail of the linked list
else {
while (tail != null) {
if (tail.next == null) {
tail.next = temp;
tail = tail.next;
}
tail = tail.next;
}
}
// Return the list
return head;
}
// Function to concatenate
// the two lists
function mergelists() {
tail = head1;
// Iterate through the
// head1 to find the
// last node join the
// next of last node
// of head1 to the
// 1st node of head2
while (tail != null) {
if (tail.next == null && head2 != null) {
tail.next = head2;
break;
}
tail = tail.next;
}
// return the concatenated
// lists as a single list - head1
return head1;
}
// Sort the linked list
// using bubble sort
function sortlist() {
curr = head1;
temp = head1;
// Compares two adjacent elements
// and swaps if the first element
// is greater than the other one.
while (curr.next != null) {
temp = curr.next;
while (temp != null) {
if (temp.data < curr.data) {
var t = temp.data;
temp.data = curr.data;
curr.data = t;
}
temp = temp.next;
}
curr = curr.next;
}
}
// Driver Code
// Given Linked List 1
head1 = getData(head1, 4);
head1 = getData(head1, 7);
head1 = getData(head1, 5);
// Given Linked List 2
head2 = getData(head2, 2);
head2 = getData(head2, 1);
head2 = getData(head2, 8);
head2 = getData(head2, 1);
// Merge the two lists
// in a single list
head1 = mergelists();
// Sort the unsorted merged list
sortlist();
// Print the final
// sorted merged list
setData(head1);
// This code is contributed by umadevi9616.
</script>
Output: 1 -> 1 -> 2 -> 4 -> 5 -> 7 -> 8
Time Complexity: O((M+N)^2) where M and N are the lengths of the two given linked lists. We are merging the two list and performing bubble sort on the merged list. The time complexity of bubble sort is quadratic.
Auxiliary Space: O(1)
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