Java Program For Union And Intersection Of Two Linked Lists
Given two Linked Lists, create union and intersection lists that contain union and intersection of the elements present in the given lists. The order of elements in output lists doesn’t matter.
Example:
Input: List1: 10->15->4->20 List2: 8->4->2->10 Output: Intersection List: 4->10 Union List: 2->8->20->4->15->10
Method 1 (Simple):
The following are simple algorithms to get union and intersection lists respectively.
1. Intersection (list1, list2):
Initialize the result list as NULL. Traverse list1 and look for every element in list2, if the element is present in list2, then add the element to the result.
2. Union (list1, list2):
Initialize the result list as NULL. Traverse list1 and add all of its elements to the result.
Traverse list2. If an element of list2 is already present in the result then do not insert it to the result, otherwise insert.
This method assumes that there are no duplicates in the given lists.
Thanks to Shekhu for suggesting this method. Following are C and Java implementations of this method.
Java
// Java program to find union and// intersection of two unsorted// linked listsclass LinkedList{ // head of list Node head; // Linked list Node class Node { int data; Node next; Node(int d) { data = d; next = null; } } /* Function to get Union of 2 Linked Lists */ void getUnion(Node head1, Node head2) { Node t1 = head1, t2 = head2; // Insert all elements of list1 // in the result while (t1 != null) { push(t1.data); t1 = t1.next; } // Insert those elements of list2 // that are not present while (t2 != null) { if (!isPresent(head, t2.data)) push(t2.data); t2 = t2.next; } } void getIntersection(Node head1, Node head2) { Node result = null; Node t1 = head1; // Traverse list1 and search each // element of it in list2. // If the element is present in // list 2, then insert the // element to result while (t1 != null) { if (isPresent(head2, t1.data)) push(t1.data); t1 = t1.next; } } // Utility function to print list void printList() { Node temp = head; while (temp != null) { System.out.print(temp.data + " "); temp = temp.next; } System.out.println(); } /* Inserts a node at start of linked list */ void push(int new_data) { /* 1 & 2: Allocate the Node & Put in the data*/ Node new_node = new Node(new_data); /* 3. Make next of new Node as head */ new_node.next = head; /* 4. Move the head to point to new Node */ head = new_node; } /* A utility function that returns true if data is present in linked list else return false */ boolean isPresent(Node head, int data) { Node t = head; while (t != null) { if (t.data == data) return true; t = t.next; } return false; } // Driver code public static void main(String args[]) { LinkedList llist1 = new LinkedList(); LinkedList llist2 = new LinkedList(); LinkedList unin = new LinkedList(); LinkedList intersection = new LinkedList(); /* Create a linked lists 10->15->5->20 */ llist1.push(20); llist1.push(4); llist1.push(15); llist1.push(10); /* Create a linked lists 8->4->2->10 */ llist2.push(10); llist2.push(2); llist2.push(4); llist2.push(8); intersection.getIntersection(llist1.head, llist2.head); unin.getUnion(llist1.head, llist2.head); System.out.println("First List is"); llist1.printList(); System.out.println("Second List is"); llist2.printList(); System.out.println("Intersection List is"); intersection.printList(); System.out.println("Union List is"); unin.printList(); }}// This code is contributed by Rajat Mishra |
Output:
First list is 10 15 4 20 Second list is 8 4 2 10 Intersection list is 4 10 Union list is 2 8 20 4 15 10
Complexity Analysis:
- Time Complexity: O(m*n).
Here ‘m’ and ‘n’ are number of elements present in the first and second lists respectively.
For union: For every element in list-2 we check if that element is already present in the resultant list made using list-1.
For intersection: For every element in list-1 we check if that element is also present in list-2. - Auxiliary Space: O(1).
No use of any data structure for storing values.
Method 2 (Use Merge Sort):
In this method, algorithms for Union and Intersection are very similar. First, we sort the given lists, then we traverse the sorted lists to get union and intersection.
The following are the steps to be followed to get union and intersection lists.
- Sort the first Linked List using merge sort. This step takes O(mLogm) time. Refer this post for details of this step.
- Sort the second Linked List using merge sort. This step takes O(nLogn) time. Refer this post for details of this step.
- Linearly scan both sorted lists to get the union and intersection. This step takes O(m + n) time. This step can be implemented using the same algorithm as sorted arrays algorithm discussed here.
The time complexity of this method is O(mLogm + nLogn) which is better than method 1’s time complexity.
Method 3 (Use Hashing):
1. Union (list1, list2):
Initialize the result list as NULL and create an empty hash table. Traverse both lists one by one, for each element being visited, look at the element in the hash table. If the element is not present, then insert the element into the result list. If the element is present, then ignore it.
2. Intersection (list1, list2)
Initialize the result list as NULL and create an empty hash table. Traverse list1. For each element being visited in list1, insert the element in the hash table. Traverse list2, for each element being visited in list2, look the element in the hash table. If the element is present, then insert the element to the result list. If the element is not present, then ignore it.
Both of the above methods assume that there are no duplicates.
Java
// Java code for Union and Intersection// of two Linked Listsimport java.util.HashMap;import java.util.HashSet;class LinkedList{ // head of list Node head; // Linked list Node class Node { int data; Node next; Node(int d) { data = d; next = null; } } // Utility function to print list void printList() { Node temp = head; while (temp != null) { System.out.print(temp.data + " "); temp = temp.next; } System.out.println(); } /* Inserts a node at start of linked list */ void push(int new_data) { /* 1 & 2: Allocate the Node & Put in the data*/ Node new_node = new Node(new_data); /* 3. Make next of new Node as head */ new_node.next = head; /* 4. Move the head to point to new Node */ head = new_node; } public void append(int new_data) { if (this.head == null) { Node n = new Node(new_data); this.head = n; return; } Node n1 = this.head; Node n2 = new Node(new_data); while (n1.next != null) { n1 = n1.next; } n1.next = n2; n2.next = null; } /* A utility function that returns true if data is present in linked list else return false */ boolean isPresent(Node head, int data) { Node t = head; while (t != null) { if (t.data == data) return true; t = t.next; } return false; } LinkedList getIntersection(Node head1, Node head2) { HashSet<Integer> hset = new HashSet<>(); Node n1 = head1; Node n2 = head2; LinkedList result = new LinkedList(); // Loop stores all the elements of // list1 in hset while (n1 != null) { if (hset.contains(n1.data)) { hset.add(n1.data); } else { hset.add(n1.data); } n1 = n1.next; } // For every element of list2 present // in hset loop inserts the element // into the result while (n2 != null) { if (hset.contains(n2.data)) { result.push(n2.data); } n2 = n2.next; } return result; } LinkedList getUnion(Node head1, Node head2) { // HashMap that will store the // elements of the lists with their counts HashMap<Integer, Integer> hmap = new HashMap<>(); Node n1 = head1; Node n2 = head2; LinkedList result = new LinkedList(); // Loop inserts the elements and the // count of that element of list1 into // the hmap while (n1 != null) { if (hmap.containsKey(n1.data)) { int val = hmap.get(n1.data); hmap.put(n1.data, val + 1); } else { hmap.put(n1.data, 1); } n1 = n1.next; } // Loop further adds the elements of // list2 with their counts into the hmap while (n2 != null) { if (hmap.containsKey(n2.data)) { int val = hmap.get(n2.data); hmap.put(n2.data, val + 1); } else { hmap.put(n2.data, 1); } n2 = n2.next; } // Eventually add all the elements // into the result that are present in the hmap for (int a : hmap.keySet()) { result.append(a); } return result; } // Driver code public static void main(String args[]) { LinkedList llist1 = new LinkedList(); LinkedList llist2 = new LinkedList(); LinkedList union = new LinkedList(); LinkedList intersection = new LinkedList(); /*create a linked list 10->15->4->20 */ llist1.push(20); llist1.push(4); llist1.push(15); llist1.push(10); /*create a linked list 8->4->2->10 */ llist2.push(10); llist2.push(2); llist2.push(4); llist2.push(8); intersection = intersection.getIntersection(llist1.head, llist2.head); union = union.getUnion(llist1.head, llist2.head); System.out.println("First List is"); llist1.printList(); System.out.println("Second List is"); llist2.printList(); System.out.println("Intersection List is"); intersection.printList(); System.out.println("Union List is"); union.printList(); }}// This code is contributed by Kamal Rawal |
Output:
First List is 10 15 4 20 Second List is 8 4 2 10 Intersection List is 10 4 Union List is 2 4 20 8 10 15
Complexity Analysis:
- Time Complexity: O(m+n).
Here ‘m’ and ‘n’ are number of elements present in the first and second lists respectively.
For union: Traverse both the lists, store the elements in Hash-map and update the respective count.
For intersection: First traverse list-1, store its elements in Hash-map and then for every element in list-2 check if it is already present in the map. This takes O(1) time. - Auxiliary Space:O(m+n).
Use of Hash-map data structure for storing values.
Please refer complete article on Union and Intersection of two Linked Lists for more details!
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