Given two linked lists, your task is to complete the function make union(), which returns the union of two linked lists. This union should include all the distinct elements only. The new list formed should be in non-decreasing order.
Input: L1 = 9->6->4->2->3->8
L2 = 1->2->8->6->2
Output: 1 2 3 4 6 8 9
Approach:
There are two types of heap as we all know that is min heap and max heap
- Min heap – Stores all the elements in ascending order
- Max heap – Stores all the elements in descending order
Let us first visualize using min heap in order to interpret program execution how union of two linked list is carried on, so de we are having two operations as listed to visualize:
- Insertion
- Removal
Insertion: After inserting all the distinct elements of two linked lists,
Removal: Removing the root until minheap is empty
Removing the root with value 1:
Removing the root with value 2:
Removing the root with value 3:
Removing the root with value 4:
Removing the root with value 6:
Removing the root with value 8:
Removing the root with value 9:
Hence we can conclude that in the min-heap, the smallest element will be at the root of the heap, and in the max heap, the greatest element will be at the root of the heap. While implementing the remove() function on heap, the root element will be removed. Since the output should be in increasing order, Min heap can be used. Priority Queue is used to implement min heap in java.
Example
Java
// JAva Program toIllustrate Union of Two Linked Lists// Using Priority Queue// Importing basic required classesimport java.io.*;import java.util.*;// Class 1// Helper class// Node creationclass Node { // Data and addressing variable of node int data; Node next; // Constructor to initialize node Node(int a) { data = a; next = null; }}// Class 2// Main classpublic class GfG { // Reading input via Scanner class static Scanner sc = new Scanner(System.in); // Method 1 // To create the input list1 public static Node inputList1() { // Declaring node variables that is // Head and tail Node head, tail; // Custom input node elements head = tail = new Node(9); tail.next = new Node(6); // Fetching for next node // using next() method tail = tail.next; // Similarly for node 3 tail.next = new Node(4); tail = tail.next; // Similarly for node 4 tail.next = new Node(2); tail = tail.next; // Similarly for node 5 tail.next = new Node(3); tail = tail.next; // Similarly for node 6 tail.next = new Node(8); tail = tail.next; // Returning the head return head; } // Method 2 // To create the input List2 // Similar to method 1 but for List2 public static Node inputList2() { Node head, tail; head = tail = new Node(1); tail.next = new Node(2); tail = tail.next; tail.next = new Node(8); tail = tail.next; tail.next = new Node(6); tail = tail.next; tail.next = new Node(2); tail = tail.next; return head; } // Method 3 // To print the union list public static void printList(Node n) { // Till there is a node // condition holds true while (n != null) { // Print the node System.out.print(n.data + " "); // Moving onto next node n = n.next; } } // Method 4 // main driver method public static void main(String args[]) { // Taking input for List 1 and List 2 Node head1 = inputList1(); Node head2 = inputList2(); // Calling Union obj = new Union(); printList(obj.findUnion(head1, head2)); }}// Class 3// To make the union of two linked listclass Union { public static Node findUnion(Node head1, Node head2) { // Creating a priority queue where // declaring elements of integer type PriorityQueue<Integer> minheap = new PriorityQueue<Integer>(); // Setting heads Node l1 = head1, l2 = head2; // For List 1 // Inserting elements from linked list1 into // priority queue while (l1 != null) { if (!minheap.contains(l1.data)) { minheap.add(l1.data); } // Move to next element l1 = l1.next; } // For List 2 // Inserting elements from linked list2 into // priority queue while (l2 != null) { if (!minheap.contains(l2.data)) { minheap.add(l2.data); } // Move to next element l2 = l2.next; } Node union = new Node(0), start = union; // Removing until heap is empty while (!minheap.isEmpty()) { Node temp = new Node(minheap.remove()); // Using temp to store start start.next = temp; start = start.next; } // Returning node return union.next; }} |
1 2 3 4 6 8 9
Time complexity: O(nlogn), Space complexity: O(n)
