Find count of common nodes in two Doubly Linked Lists
Given two doubly linked lists. The task is to find the total number of common nodes in both the doubly linked list.
Examples:
Input : list 1 = 15 <=> 16 <=> 10 <=> 9 <=> 6 <=> 7 <=> 17 list 2 = 15 <=> 16 <=> 45 <=> 9 <=> 6 Output : Number of common nodes: 4 Input : list 1 = 18 <=> 30 <=> 92 <=> 46 <=> 72 <=> 1 list 2 = 12 <=> 32 <=> 45 <=> 9 <=> 6 <=> 30 Output : Number of common nodes: 1
Approach: Traverse both lists till the end of the list using two nested loops. For every node in list 1 check if it matches with any node in list 2. If yes then increment the count of common nodes. Finally, print the count.
Below is the implementation of the above approach:
C++
// C++ implementation to count// common element in given two// doubly linked list#include <bits/stdc++.h>using namespace std;// Node of the doubly linked liststruct Node { int data; Node *prev, *next;};// Function to insert a node at the beginning// of the Doubly Linked Listvoid push(Node** head_ref, int new_data){ // allocate node Node* new_node = (Node*)malloc(sizeof(struct Node)); // put in the data new_node->data = new_data; // since we are adding at the beginning, // prev is always NULL new_node->prev = NULL; // link the old list off the new node new_node->next = (*head_ref); // change prev of head node to new node if ((*head_ref) != NULL) (*head_ref)->prev = new_node; // move the head to point to the new node (*head_ref) = new_node;}// Count common nodes in both list1 and list 2int countCommonNodes(Node** head_ref, Node** head){ // head for list 1 Node* ptr = *head_ref; // head for list 2 Node* ptr1 = *head; // initialize count = 0 int count = 0; // traverse list 1 till the end while (ptr != NULL) { // traverse list 2 till the end while (ptr1 != NULL) { // if node value is equal then // increment count if (ptr->data == ptr1->data) { count++; break; } // increment pointer list 2 ptr1 = ptr1->next; } // again list 2 start with starting point ptr1 = *head; // increment pointer list 1 ptr = ptr->next; } // return count of common nodes return count;}// Driver programint main(){ // start with the empty list Node* head = NULL; Node* head1 = NULL; // create the doubly linked list 1 // 15 <-> 16 <-> 10 <-> 9 <-> 6 <-> 7 <-> 17 push(&head, 17); push(&head, 7); push(&head, 6); push(&head, 9); push(&head, 10); push(&head, 16); push(&head, 15); // create the doubly linked list 2 // 15 <-> 16 <-> 45 <-> 9 <-> 6 push(&head1, 6); push(&head1, 9); push(&head1, 45); push(&head1, 16); push(&head1, 15); cout << "Number of common nodes:" << countCommonNodes(&head, &head1); return 0;} |
Java
// Java implementation to count// common element in given two// doubly linked listclass GFG{// Node of the doubly linked liststatic class Node{ int data; Node prev, next;};// Function to insert a node at the beginning// of the Doubly Linked Liststatic Node push(Node head_ref, int new_data){ // allocate node Node new_node = new Node(); // put in the data new_node.data = new_data; // since we are adding at the beginning, // prev is always null new_node.prev = null; // link the old list off the new node new_node.next = head_ref; // change prev of head node to new node if (head_ref != null) head_ref.prev = new_node; // move the head to point to the new node head_ref = new_node; return head_ref;}// Count common nodes in both list1 and list 2static int countCommonNodes(Node head_ref, Node head){ // head for list 1 Node ptr = head_ref; // head for list 2 Node ptr1 = head; // initialize count = 0 int count = 0; // traverse list 1 till the end while (ptr != null) { // traverse list 2 till the end while (ptr1 != null) { // if node value is equal then // increment count if (ptr.data == ptr1.data) { count++; break; } // increment pointer list 2 ptr1 = ptr1.next; } // again list 2 start with starting point ptr1 = head; // increment pointer list 1 ptr = ptr.next; } // return count of common nodes return count;}// Driver Codepublic static void main(String[] args){ // start with the empty list Node head = null; Node head1 = null; // create the doubly linked list 1 // 15 <. 16 <. 10 <. 9 <. 6 <. 7 <. 17 head = push(head, 17); head = push(head, 7); head = push(head, 6); head = push(head, 9); head = push(head, 10); head = push(head, 16); head = push(head, 15); // create the doubly linked list 2 // 15 <. 16 <. 45 <. 9 <. 6 head1 = push(head1, 6); head1 = push(head1, 9); head1 = push(head1, 45); head1 = push(head1, 16); head1 = push(head1, 15); System.out.println("Number of common nodes: " + countCommonNodes(head, head1));}}// This code is contributed by Rajput-Ji |
Python3
# Python3 implementation to count# common element in given two# doubly linked list # Node of the doubly linked listclass Node: def __init__(self, data): self.data = data self.prev = None self.next = None # Function to insert a node at the beginning# of the Doubly Linked Listdef push(head_ref, new_data): # allocate node new_node = Node(new_data) # put in the data new_node.data = new_data; # since we are adding at the beginning, # prev is always None new_node.prev = None; # link the old list off the new node new_node.next = (head_ref); # change prev of head node to new node if ((head_ref) != None): (head_ref).prev = new_node; # move the head to point to the new node (head_ref) = new_node; return head_ref# Count common nodes in both list1 and list 2def countCommonNodes(head_ref, head): # head for list 1 ptr = head_ref; # head for list 2 ptr1 = head; # initialize count = 0 count = 0; # traverse list 1 till the end while (ptr != None): # traverse list 2 till the end while (ptr1 != None): # if node value is equal then # increment count if (ptr.data == ptr1.data): count += 1 break; # increment pointer list 2 ptr1 = ptr1.next; # again list 2 start with starting point ptr1 = head; # increment pointer list 1 ptr = ptr.next; # return count of common nodes return count; # Driver programif __name__=='__main__': # start with the empty list head = None; head1 = None; # create the doubly linked list 1 # 15 <. 16 <. 10 <. 9 <. 6 <. 7 <. 17 head = push(head, 17); head = push(head, 7); head = push(head, 6); head = push(head, 9); head = push(head, 10); head = push(head, 16); head = push( head, 15); # create the doubly linked list 2 # 15 <. 16 <. 45 <. 9 <. 6 head1 = push(head1, 6); head1 = push(head1, 9); head1 = push(head1, 45); head1 = push(head1, 16); head1 = push(head1, 15); print("Number of common nodes: " + str(countCommonNodes(head, head1))) # This code is contributed by rutvik_56. |
C#
// C# implementation to count// common element in given two// doubly linked listusing System; class GFG{// Node of the doubly linked listpublic class Node{ public int data; public Node prev, next;};// Function to insert a node at the beginning// of the Doubly Linked Liststatic Node push(Node head_ref, int new_data){ // allocate node Node new_node = new Node(); // put in the data new_node.data = new_data; // since we are adding at the beginning, // prev is always null new_node.prev = null; // link the old list off the new node new_node.next = head_ref; // change prev of head node to new node if (head_ref != null) head_ref.prev = new_node; // move the head to point to the new node head_ref = new_node; return head_ref;}// Count common nodes in both list1 and list 2static int countCommonNodes(Node head_ref, Node head){ // head for list 1 Node ptr = head_ref; // head for list 2 Node ptr1 = head; // initialize count = 0 int count = 0; // traverse list 1 till the end while (ptr != null) { // traverse list 2 till the end while (ptr1 != null) { // if node value is equal then // increment count if (ptr.data == ptr1.data) { count++; break; } // increment pointer list 2 ptr1 = ptr1.next; } // again list 2 start with starting point ptr1 = head; // increment pointer list 1 ptr = ptr.next; } // return count of common nodes return count;}// Driver Codepublic static void Main(String[] args){ // start with the empty list Node head = null; Node head1 = null; // create the doubly linked list 1 // 15 <. 16 <. 10 <. 9 <. 6 <. 7 <. 17 head = push(head, 17); head = push(head, 7); head = push(head, 6); head = push(head, 9); head = push(head, 10); head = push(head, 16); head = push(head, 15); // create the doubly linked list 2 // 15 <. 16 <. 45 <. 9 <. 6 head1 = push(head1, 6); head1 = push(head1, 9); head1 = push(head1, 45); head1 = push(head1, 16); head1 = push(head1, 15); Console.WriteLine("Number of common nodes: " + countCommonNodes(head, head1));}}// This code is contributed by PrinciRaj1992 |
Javascript
<script>// JavaScript implementation to count// common element in given two// doubly linked list // Node of the doubly linked listclass Node { constructor(val) { this.data = val; this.prev = null; this.next = null; }} // Function to insert a node at the beginning // of the Doubly Linked List function push(head_ref , new_data) { // allocate node var new_node = new Node(); // put in the data new_node.data = new_data; // since we are adding at the beginning, // prev is always null new_node.prev = null; // link the old list off the new node new_node.next = head_ref; // change prev of head node to new node if (head_ref != null) head_ref.prev = new_node; // move the head to povar to the new node head_ref = new_node; return head_ref; } // Count common nodes in both list1 and list 2 function countCommonNodes(head_ref, head) { // head for list 1 var ptr = head_ref; // head for list 2 var ptr1 = head; // initialize count = 0 var count = 0; // traverse list 1 till the end while (ptr != null) { // traverse list 2 till the end while (ptr1 != null) { // if node value is equal then // increment count if (ptr.data == ptr1.data) { count++; break; } // increment pointer list 2 ptr1 = ptr1.next; } // again list 2 start with starting point ptr1 = head; // increment pointer list 1 ptr = ptr.next; } // return count of common nodes return count; } // Driver Code // start with the empty list var head = null; var head1 = null; // create the doubly linked list 1 // 15 <. 16 <. 10 <. 9 <. 6 <. 7 <. 17 head = push(head, 17); head = push(head, 7); head = push(head, 6); head = push(head, 9); head = push(head, 10); head = push(head, 16); head = push(head, 15); // create the doubly linked list 2 // 15 <. 16 <. 45 <. 9 <. 6 head1 = push(head1, 6); head1 = push(head1, 9); head1 = push(head1, 45); head1 = push(head1, 16); head1 = push(head1, 15); document.write("Number of common nodes: " + countCommonNodes(head, head1));// This code contributed by umadevi9616</script> |
Output:
Number of common nodes:4
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