# 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:

 `// C++ program for the above approach ` `#include ` `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; ` `} `

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. My Personal Notes arrow_drop_up

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