# Union and Intersection of two linked lists | Set-2 (Using Merge Sort)

Given two Linked Lists, create union and intersection lists that contain union and intersection of the elements present in the given lists. Order of elements in output lists doesnâ€™t matter.

**Examples:**

Input:List1: 10 -> 15 -> 4 -> 20 List2: 8 -> 4 -> 2 -> 10Output:Intersection List: 4 -> 10 Union List: 2 -> 8 -> 20 -> 4 -> 15 -> 10Explanation:In this two lists 4 and 10 nodes are common. The union lists contains all the nodes of both the lists.Input:List1: 1 -> 2 -> 3 -> 4 List2: 3 -> 4 -> 8 -> 10Output:Intersection List: 3 -> 4 Union List: 1 -> 2 -> 3 -> 4 -> 8 -> 10Explanation:In this two lists 4 and 3 nodes are common. The union lists contains all the nodes of both the lists.

There were three methods discussed in this post with an implementation of Method 1. In this post, we will see an implementation of Method 2 i.e. Using Merge sort.

Implementation:Following are the steps to be followed to get union and intersection lists. 1) Sort both Linked Lists using merge sort. This step takes O(mLogm) time. 2) Linearly scan both sorted lists to get the union and intersection. This step takes O(m + n) time.

Algorithm:

- Sort both Linked Lists using merge sort.
- Linearly scan both sorted lists to get the union and intersection.
- Perform merge like operation on both linked lists, Keep to a pointer which points initially to the first node of both lists.
- Compare both the nodes until and unless both the pointers are not null.

- if equal add it to the intersection list and union list and move to next node of both the pointers
- if not equal then insert the smaller pointer value into union list and move to the next node
- If one of the pointers is null then traverse the other list and all its nodes to union list.

**Just like Method 1, this method also assumes that there are distinct elements in the lists.**

## CPP

`// C++ program to find union and intersection of` `// two unsorted linked lists in O(n Log n) time.` `#include <bits/stdc++.h>` `using` `namespace` `std;` `/* Link list node */` `struct` `Node {` ` ` `int` `data;` ` ` `struct` `Node* next;` `};` `/* A utility function to insert a` `node at the beginning of a linked list*/` `void` `push(` `struct` `Node** head_ref, ` `int` `new_data)` `{` ` ` `/* allocate node */` ` ` `struct` `Node* new_node` ` ` `= (` `struct` `Node*)` `malloc` `(` `sizeof` `(` `struct` `Node));` ` ` `/* put in the data */` ` ` `new_node->data = new_data;` ` ` `/* link the old list off the new node */` ` ` `new_node->next = (*head_ref);` ` ` `/* move the head to point to the new node */` ` ` `(*head_ref) = new_node;` `}` `/* UTILITY FUNCTIONS */` `/* Split the nodes of the given` `list into front and back halves,` `and return the two lists` `using the reference parameters.` `If the length is odd, the` `extra node should go in the` `front list.` ` ` `Uses the fast/slow pointer strategy. */` `void` `FrontBackSplit(` `struct` `Node* source,` ` ` `struct` `Node** frontRef,` ` ` `struct` `Node** backRef)` `{` ` ` `struct` `Node* fast;` ` ` `struct` `Node* slow;` ` ` `if` `(source == NULL || source->next == NULL) {` ` ` `/* length < 2 cases */` ` ` `*frontRef = source;` ` ` `*backRef = NULL;` ` ` `}` ` ` `else` `{` ` ` `slow = source;` ` ` `fast = source->next;` ` ` `/* Advance 'fast' two nodes, and` ` ` `advance 'slow' one node */` ` ` `while` `(fast != NULL) {` ` ` `fast = fast->next;` ` ` `if` `(fast != NULL) {` ` ` `slow = slow->next;` ` ` `fast = fast->next;` ` ` `}` ` ` `}` ` ` `/* 'slow' is before the midpoint in the list,` ` ` `so split it in two at that point. */` ` ` `*frontRef = source;` ` ` `*backRef = slow->next;` ` ` `slow->next = NULL;` ` ` `}` `}` `/* See https:// www.geeksforgeeks.org/?p=3622 for details` `of this function */` `struct` `Node* SortedMerge(` `struct` `Node* a, ` `struct` `Node* b)` `{` ` ` `struct` `Node* result = NULL;` ` ` `/* Base cases */` ` ` `if` `(a == NULL)` ` ` `return` `(b);` ` ` `else` `if` `(b == NULL)` ` ` `return` `(a);` ` ` `/* Pick either a or b, and recur */` ` ` `if` `(a->data <= b->data) {` ` ` `result = a;` ` ` `result->next = SortedMerge(a->next, b);` ` ` `}` ` ` `else` `{` ` ` `result = b;` ` ` `result->next = SortedMerge(a, b->next);` ` ` `}` ` ` `return` `(result);` `}` `/* sorts the linked list by changing` `next pointers (not data) */` `void` `mergeSort(` `struct` `Node** headRef)` `{` ` ` `struct` `Node* head = *headRef;` ` ` `struct` `Node *a, *b;` ` ` `/* Base case -- length 0 or 1 */` ` ` `if` `((head == NULL) || (head->next == NULL))` ` ` `return` `;` ` ` `/* Split head into 'a' and 'b' sublists */` ` ` `FrontBackSplit(head, &a, &b);` ` ` `/* Recursively sort the sublists */` ` ` `mergeSort(&a);` ` ` `mergeSort(&b);` ` ` `/* answer = merge the two sorted lists together */` ` ` `*headRef = SortedMerge(a, b);` `}` `/* Function to get union of two` `linked lists head1 and head2 */` `struct` `Node* getUnion(` `struct` `Node* head1,` ` ` `struct` `Node* head2)` `{` ` ` `struct` `Node* result = NULL;` ` ` `struct` `Node *t1 = head1, *t2 = head2;` ` ` `// Traverse both lists and store the` ` ` `// element in the resu1tant list` ` ` `while` `(t1 != NULL && t2 != NULL) {` ` ` `// Move to the next of first list` ` ` `// if its element is smaller` ` ` `if` `(t1->data < t2->data) {` ` ` `push(&result, t1->data);` ` ` `t1 = t1->next;` ` ` `}` ` ` `// Else move to the next of second list` ` ` `else` `if` `(t1->data > t2->data) {` ` ` `push(&result, t2->data);` ` ` `t2 = t2->next;` ` ` `}` ` ` `// If same then move to the next node` ` ` `// in both lists` ` ` `else` `{` ` ` `push(&result, t2->data);` ` ` `t1 = t1->next;` ` ` `t2 = t2->next;` ` ` `}` ` ` `}` ` ` `/* Print remaining elements of the lists */` ` ` `while` `(t1 != NULL) {` ` ` `push(&result, t1->data);` ` ` `t1 = t1->next;` ` ` `}` ` ` `while` `(t2 != NULL) {` ` ` `push(&result, t2->data);` ` ` `t2 = t2->next;` ` ` `}` ` ` `return` `result;` `}` `/* Function to get intersection of` `two linked lists head1 and head2 */` `struct` `Node* getIntersection(` `struct` `Node* head1,` ` ` `struct` `Node* head2)` `{` ` ` `struct` `Node* result = NULL;` ` ` `struct` `Node *t1 = head1, *t2 = head2;` ` ` `// Traverse both lists and store the same element` ` ` `// in the resu1tant list` ` ` `while` `(t1 != NULL && t2 != NULL) {` ` ` `// Move to the next of first` ` ` `// list if smaller` ` ` `if` `(t1->data < t2->data)` ` ` `t1 = t1->next;` ` ` `// Move to the next of second` ` ` `// list if it is smaller` ` ` `else` `if` `(t1->data > t2->data)` ` ` `t2 = t2->next;` ` ` `// If both are same` ` ` `else` `{` ` ` `// Store current element in the list` ` ` `push(&result, t2->data);` ` ` `// Move to the next node of both lists` ` ` `t1 = t1->next;` ` ` `t2 = t2->next;` ` ` `}` ` ` `}` ` ` `// return the resultant list` ` ` `return` `result;` `}` `/* A utility function to print a linked list*/` `void` `printList(` `struct` `Node* node)` `{` ` ` `while` `(node != NULL) {` ` ` `printf` `(` `"%d "` `, node->data);` ` ` `node = node->next;` ` ` `}` `}` `/* Driver program to test above function*/` `int` `main()` `{` ` ` `/* Start with the empty list */` ` ` `struct` `Node* head1 = NULL;` ` ` `struct` `Node* head2 = NULL;` ` ` `struct` `Node* intersection_list = NULL;` ` ` `struct` `Node* union_list = NULL;` ` ` `/*create a linked list 11->10->15->4->20 */` ` ` `push(&head1, 20);` ` ` `push(&head1, 4);` ` ` `push(&head1, 15);` ` ` `push(&head1, 10);` ` ` `push(&head1, 11);` ` ` `/*create a linked list 8->4->2->10 */` ` ` `push(&head2, 10);` ` ` `push(&head2, 2);` ` ` `push(&head2, 4);` ` ` `push(&head2, 8);` ` ` `/* Sort the above created Linked List */` ` ` `mergeSort(&head1);` ` ` `mergeSort(&head2);` ` ` `intersection_list = getIntersection(head1, head2);` ` ` `union_list = getUnion(head1, head2);` ` ` `printf` `(` `"First list is \n"` `);` ` ` `printList(head1);` ` ` `printf` `(` `"\nSecond list is \n"` `);` ` ` `printList(head2);` ` ` `printf` `(` `"\nIntersection list is \n"` `);` ` ` `printList(intersection_list);` ` ` `printf` `(` `"\nUnion list is \n"` `);` ` ` `printList(union_list);` ` ` `return` `0;` `}` |

**Output**

First list is 4 10 11 15 20 Second list is 2 4 8 10 Intersection list is 10 4 Union list is 20 15 11 10 8 4 2

**Complexity Analysis:**

**Time complexity:**O(m Log m + n Log n). Time required to sort the lists are n log n and m log m and to find union and intersection linear time is required.**Auxiliary Space:**O(m+n). If the output is stored then O(m+n) space is required.

In the next post, Method-3 will be discussed i.e. using hashing. This article is contributed by **Sahil Chhabra**. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.