Merge two unsorted linked lists to get a sorted list

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 its 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 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:

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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;
}

chevron_right


Output:

1 -> 1 -> 2 -> 4 -> 5 -> 7 -> 8

Time Complexity: O(M*N) where M and N are the length of the two given linked lists.

competitive-programming-img




My Personal Notes arrow_drop_up

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.