Merge two unsorted linked lists to get a sorted list (original) (raw)

Last Updated : 21 Aug, 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:

1. 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.
2. 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

// 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

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#

// 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

JavaScript

`

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)