C++ Program For Union And Intersection Of Two Linked Lists
Last Updated :
12 Oct, 2023
Write a C++ program for a 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.
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.
Union (list1, list2):
Initialize a new list ans and store first and second list data to set to remove duplicate data
and then store it into our new list ans and return its head.
Below is the implementation of above approach:
C++
// C++ program to find union
// and intersection of two unsorted
// linked lists
#include "bits/stdc++.h"
using namespace std;
/* Linked list node */
struct Node {
int data;
struct Node* next;
Node(int x)
{
data = x;
next = NULL;
}
};
/* A utility function to insert a
node at the beginning ofa linked list*/
void push(struct Node** head_ref, int new_data);
/* A utility function to check if
given data is present in a list */
bool isPresent(struct Node* head, int data);
/* Function to get union of two
linked lists head1 and head2 */
struct Node* getUnion(struct Node* head1,
struct Node* head2)
{
struct Node* ans = new Node(-1);
struct Node* head = ans;
set<int> st;
while (head1 != NULL) {
st.insert(head1->data);
head1 = head1->next;
}
while (head2 != NULL) {
st.insert(head2->data);
head2 = head2->next;
}
for (auto it : st) {
struct Node* t = new Node(it);
ans->next = t;
ans = ans->next;
}
head = head->next;
return head;
}
/* 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;
// 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(&result, t1->data);
t1 = t1->next;
}
return result;
}
/* 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 of the new node */
new_node->next = (*head_ref);
/* move the head to point to the new node */
(*head_ref) = new_node;
}
/* A utility function to print a linked list*/
void printList(struct Node* node)
{
while (node != NULL) {
cout << " " << node->data;
node = node->next;
}
}
bool isPresent(struct Node* head, int data)
{
struct Node* t = head;
while (t != NULL) {
if (t->data == data)
return 1;
t = t->next;
}
return 0;
}
/* Driver program to test above function*/
int main()
{
/* Start with the empty list */
struct Node* head1 = NULL;
struct Node* head2 = NULL;
struct Node* intersecn = NULL;
struct Node* unin = NULL;
/*create a linked lists 10->15->5->20 */
push(&head1, 20);
push(&head1, 4);
push(&head1, 15);
push(&head1, 10);
/*create a linked lists 8->4->2->10 */
push(&head2, 10);
push(&head2, 2);
push(&head2, 4);
push(&head2, 8);
intersecn = getIntersection(head1, head2);
unin = getUnion(head1, head2);
cout << "\n First list is " << endl;
printList(head1);
cout << "\n Second list is " << endl;
printList(head2);
cout << "\n Intersection list is " << endl;
printList(intersecn);
cout << "\n Union list is " << endl;
printList(unin);
return 0;
}
// This code is contributed by zishanahmad786
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.
Below is the implementation of above approach:
C++
#include <iostream>
using namespace std;
class Node {
public:
int data;
Node* next;
Node(int data)
{
this->data = data;
this->next = NULL;
}
};
// function to print linked list
void printLinkedList(Node* head)
{
Node* temp = head;
while (temp != NULL) {
cout << temp->data << "-->";
temp = temp->next;
}
cout << "None";
}
// function to get union of two linked lists
Node* getUnion(Node* ll1, Node* ll2)
{
Node* tail = NULL;
Node* head = NULL;
while (ll1 != NULL || ll2 != NULL) {
if (ll1 == NULL) {
if (tail == NULL) {
tail = new Node(ll2->data);
head = tail;
}
else {
tail->next = new Node(ll2->data);
tail = tail->next;
}
ll2 = ll2->next;
}
else if (ll2 == NULL) {
if (tail == NULL) {
tail = new Node(ll1->data);
head = tail;
}
else {
tail->next = new Node(ll1->data);
tail = tail->next;
}
ll1 = ll1->next;
}
else {
if (ll1->data < ll2->data) {
if (tail == NULL) {
tail = new Node(ll1->data);
head = tail;
}
else {
tail->next = new Node(ll1->data);
tail = tail->next;
}
ll1 = ll1->next;
}
else if (ll1->data > ll2->data) {
if (tail == NULL) {
tail = new Node(ll2->data);
head = tail;
}
else {
tail->next = new Node(ll2->data);
tail = tail->next;
}
ll2 = ll2->next;
}
else {
if (tail == NULL) {
tail = new Node(ll1->data);
head = tail;
}
else {
tail->next = new Node(ll1->data);
tail = tail->next;
}
ll1 = ll1->next;
ll2 = ll2->next;
}
}
}
return head;
}
// main function to test the code
int main()
{
// create first linked list
Node* head1 = new Node(10);
head1->next = new Node(20);
head1->next->next = new Node(30);
head1->next->next->next = new Node(40);
head1->next->next->next->next = new Node(50);
head1->next->next->next->next->next = new Node(60);
head1->next->next->next->next->next->next
= new Node(70);
// create second linked list
Node* head2 = new Node(10);
head2->next = new Node(30);
head2->next->next = new Node(50);
head2->next->next->next = new Node(80);
head2->next->next->next->next = new Node(90);
Node* head = getUnion(head1, head2);
printLinkedList(head);
cout << endl;
return 0;
}
// This code is contributed by Gaurav
Output10-->20-->30-->40-->50-->60-->70-->80-->90-->None
Method 3 (Use Hashing):
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.
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.
Below is the implementation of above approach:
C++
#include <iostream>
#include <map>
#include <unordered_set>
using namespace std;
class LinkedList {
public:
struct Node {
int data;
Node* next;
Node(int d)
: data(d)
, next(nullptr)
{
}
};
Node* head = nullptr;
void printList()
{
Node* temp = head;
while (temp != nullptr) {
cout << temp->data << " ";
temp = temp->next;
}
cout << endl;
}
void push(int new_data)
{
Node* new_node = new Node(new_data);
new_node->next = head;
head = new_node;
}
void append(int new_data)
{
if (head == nullptr) {
Node* n = new Node(new_data);
head = n;
return;
}
Node* n1 = head;
Node* n2 = new Node(new_data);
while (n1->next != nullptr) {
n1 = n1->next;
}
n1->next = n2;
n2->next = nullptr;
}
bool isPresent(Node* head, int data)
{
Node* t = head;
while (t != nullptr) {
if (t->data == data)
return true;
t = t->next;
}
return false;
}
LinkedList getIntersection(Node* head1, Node* head2)
{
unordered_set<int> hset;
Node* n1 = head1;
Node* n2 = head2;
LinkedList result;
while (n1 != nullptr) {
if (hset.find(n1->data) == hset.end()) {
hset.insert(n1->data);
}
n1 = n1->next;
}
while (n2 != nullptr) {
if (hset.find(n2->data) != hset.end()) {
result.push(n2->data);
}
n2 = n2->next;
}
return result;
}
LinkedList getUnion(Node* head1, Node* head2)
{
map<int, int> hmap;
Node* n1 = head1;
Node* n2 = head2;
LinkedList result;
while (n1 != nullptr) {
if (hmap.find(n1->data) != hmap.end()) {
hmap[n1->data]++;
}
else {
hmap[n1->data] = 1;
}
n1 = n1->next;
}
while (n2 != nullptr) {
if (hmap.find(n2->data) != hmap.end()) {
hmap[n2->data]++;
}
else {
hmap[n2->data] = 1;
}
n2 = n2->next;
}
for (auto it = hmap.begin(); it != hmap.end();
it++) {
result.append(it->first);
}
return result;
}
};
int main()
{
LinkedList llist1, llist2, intersection, union_list;
llist1.push(20);
llist1.push(4);
llist1.push(15);
llist1.push(10);
llist2.push(10);
llist2.push(2);
llist2.push(4);
llist2.push(8);
intersection = intersection.getIntersection(
llist1.head, llist2.head);
union_list
= union_list.getUnion(llist1.head, llist2.head);
cout << "First List is" << endl;
llist1.printList();
cout << "Second List is" << endl;
llist2.printList();
cout << "Intersection List is" << endl;
intersection.printList();
cout << "Union List is" << endl;
;
union_list.printList();
}
// This code is contributed by Gaurav_Arora
OutputFirst List is
10 15 4 20
Second List is
8 4 2 10
Intersection List is
10 4
Union List is
2 4 8 10 15 20
Complexity Analysis:
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!
Similar Reads
C++ Program For Finding Intersection Point Of Two Linked Lists
There are two singly linked lists in a system. By some programming error, the end node of one of the linked lists got linked to the second list, forming an inverted Y-shaped list. Write a program to get the point where two linked lists merge. Above diagram shows an example with two linked lists hav
9 min read
C++ Program For Finding Intersection Of Two Sorted Linked Lists
Given two lists sorted in increasing order, create and return a new list representing the intersection of the two lists. The new list should be made with its own memory — the original lists should not be changed. Example: Input: First linked list: 1->2->3->4->6 Second linked list be 2-
6 min read
C++ Program For Inserting A Node In A Linked List
Inserting a node into a linked list can be done in several ways, depending on where we want to insert the new node. Here, we'll cover four common scenarios: inserting at the front of the list, after a given node, at a specific position, and at the end of the listTable of ContentInsert a Node at the
9 min read
C++ Program For Insertion Sort In A Singly Linked List
We have discussed Insertion Sort for arrays. In this article we are going to discuss Insertion Sort for linked list. Below is a simple insertion sort algorithm for a linked list. 1) Create an empty sorted (or result) list. 2) Traverse the given list, do following for every node. ......a) Insert curr
5 min read
C++ Program For Writing A Function To Delete A Linked List
Algorithm For C++:Iterate through the linked list and delete all the nodes one by one. The main point here is not to access the next of the current pointer if the current pointer is deleted. Implementation: C++ // C++ program to delete a linked list #include <bits/stdc++.h> using namespace std
2 min read
C++ Program For Writing A Function To Get Nth Node In A Linked List
Write a C++ Program to GetNth() function that takes a linked list and an integer index and returns the data value stored in the node at that index position. Example: Input: 1->10->30->14, index = 2Output: 30 The node at index 2 is 30Recommended: Please solve it on "PRACTICE" first, before m
4 min read
C++ Program For Adding Two Polynomials Using Linked List
Given two polynomial numbers represented by a linked list. Write a function that add these lists means add the coefficients who have same variable powers.Example: Input: 1st number = 5x2 + 4x1 + 2x0 2nd number = -5x1 - 5x0 Output: 5x2-1x1-3x0 Input: 1st number = 5x3 + 4x2 + 2x0 2nd number = 5x^1 - 5
5 min read
C++ Program To Check If Two Linked Lists Are Identical
Two Linked Lists are identical when they have the same data and the arrangement of data is also the same. For example, Linked lists a (1->2->3) and b(1->2->3) are identical. . Write a function to check if the given two linked lists are identical. Recommended: Please solve it on "PRACTICE
3 min read
C++ Program For Rearranging A Given Linked List In-Place
Given a singly linked list L0 -> L1 -> … -> Ln-1 -> Ln. Rearrange the nodes in the list so that the new formed list is : L0 -> Ln -> L1 -> Ln-1 -> L2 -> Ln-2 ...You are required to do this in place without altering the nodes' values. Examples: Input: 1 -> 2 -> 3 -
7 min read
Get the Intersection point of two Linked List by counting nodes
There are two singly linked lists in a system. By some programming error, the end node of one of the linked list got linked to the second list, forming an inverted Y shaped list. Write a program to get the point where two linked list merge. Above diagram shows an example with two linked list having
8 min read