Find the common nodes in two singly linked list
Given two linked list, the task is to find the number of common nodes in both singly linked list.
Examples:
Input: List A = 3 -> 4 -> 12 -> 10 -> 17, List B = 10 -> 4 -> 8 -> 575 -> 34 -> 12
Output: Number of common nodes in both list is = 3Input: List A = 12 -> 4 -> 65 -> 14 -> 59, List B = 14 -> 15 -> 23 -> 17 -> 41 -> 54
Output: Number of common nodes in both list is = 1
Naive Approach: Compare every node of list A with every node of list B. If the node is a match then increment the count and return count after all the nodes get compared.
Below is the implementation of above approach:
C++
// C++ implementation of above approach#include <bits/stdc++.h>using namespace std;// Structure of a linked list nodestruct Node { int data; struct Node* next;};// Function to common nodes which have// same value node(s) both listint countCommonNodes(struct Node* head1, struct Node* head2){ // list A struct Node* current1 = head1; // list B struct Node* current2 = head2; // set count = 0 int count = 0; // traverse list A till the end of list while (current1 != NULL) { // traverse list B till the end of list while (current2 != NULL) { // if data is match then count increase if (current1->data == current2->data) count++; // increase current pointer for next node current2 = current2->next; } // increase current pointer of list A current1 = current1->next; // initialize list B starting point current2 = head2; } // return count return count;}/* Utility function to insert a node at the beginning */void push(struct Node** head_ref, int new_data){ struct Node* new_node = new Node; new_node->data = new_data; new_node->next = *head_ref; *head_ref = new_node;}/* Utility function to print a linked list */void printList(struct Node* head){ while (head != NULL) { cout << head->data << " "; head = head->next; } cout << endl;}/* Driver program to test above functions */int main(){ struct Node* head1 = NULL; struct Node* head2 = NULL; /* Create following linked list List A = 3 -> 4 -> 12 -> 10 -> 17*/ push(&head1, 17); push(&head1, 10); push(&head1, 12); push(&head1, 4); push(&head1, 3); // List B = 10 -> 4 -> 8 -> 575 -> 34 -> 12 push(&head2, 12); push(&head2, 34); push(&head2, 575); push(&head2, 8); push(&head2, 4); push(&head2, 10); // print list A cout << "Given Linked List A: \n"; printList(head1); // print list B cout << "Given Linked List B: \n"; printList(head2); // call function for count common node int count = countCommonNodes(head1, head2); // print number of common node in both list cout << "Number of common node in both list is = " << count; return 0;} |
Java
// Java implementation of above approachclass GFG { // Structure of a linked list node static class Node { int data; Node next; }; // Function to common nodes which have // same value node(s) both list static int countCommonNodes(Node head1, Node head2) { // list A Node current1 = head1; // list B Node current2 = head2; // set count = 0 int count = 0; // traverse list A till the end of list while (current1 != null) { // traverse list B till the end of list while (current2 != null) { // if data is match then count increase if (current1.data == current2.data) count++; // increase current pointer for next node current2 = current2.next; } // increase current pointer of list A current1 = current1.next; // initialize list B starting point current2 = head2; } // return count return count; } // Utility function to insert a node at the beginning / static Node push(Node head_ref, int new_data) { Node new_node = new Node(); new_node.data = new_data; new_node.next = head_ref; head_ref = new_node; return head_ref; } // Utility function to print a linked list static void printList(Node head) { while (head != null) { System.out.print(head.data + " "); head = head.next; } System.out.println(); } // Driver code public static void main(String args[]) { Node head1 = null; Node head2 = null; // Create following linked list // List A = 3 . 4 . 12 . 10 . 17 head1 = push(head1, 17); head1 = push(head1, 10); head1 = push(head1, 12); head1 = push(head1, 4); head1 = push(head1, 3); // List B = 10 . 4 . 8 . 575 . 34 . 12 head2 = push(head2, 12); head2 = push(head2, 34); head2 = push(head2, 575); head2 = push(head2, 8); head2 = push(head2, 4); head2 = push(head2, 10); // print list A System.out.print("Given Linked List A: \n"); printList(head1); // print list B System.out.print("Given Linked List B: \n"); printList(head2); // call function for count common node int count = countCommonNodes(head1, head2); // print number of common node in both list System.out.print("Number of common node in both list is = " + count); }}// This code is contributed by Arnab Kundu |
Python3
# Python3 implementation of above approach# Link list nodeclass Node: def __init__(self, data): self.data = data self.next = None# Function to common nodes which have# same value node(s) both listdef countCommonNodes(head1, head2): # list A current1 = head1 # list B current2 = head2 # set count = 0 count = 0 # traverse list A till the end of list while (current1 != None): # traverse list B till the end of list while (current2 != None): # if data is match then count increase if (current1.data == current2.data): count = count + 1 # increase current pointer for next node current2 = current2.next # increase current pointer of list A current1 = current1.next # initialize list B starting point current2 = head2 # return count return count# Utility function to insert a node at the beginningdef push(head_ref, new_data): new_node = Node(0) new_node.data = new_data new_node.next = head_ref head_ref = new_node return head_ref# Utility function to print a linked listdef printList( head): while (head != None): print(head.data, end = " ") head = head.next print("")# Driver Codeif __name__=='__main__': head1 = None head2 = None # Create following linked list # List A = 3 . 4 . 12 . 10 . 17 head1 = push(head1, 17) head1 = push(head1, 10) head1 = push(head1, 12) head1 = push(head1, 4) head1 = push(head1, 3) # List B = 10 . 4 . 8 . 575 . 34 . 12 head2 = push(head2, 12) head2 = push(head2, 34) head2 = push(head2, 575) head2 = push(head2, 8) head2 = push(head2, 4) head2 = push(head2, 10) # print list A print("Given Linked List A: ") printList(head1) # print list B print("Given Linked List B: ") printList(head2) # call function for count common node count = countCommonNodes(head1, head2) # print number of common node in both list print("Number of common node in both list is = ", count)# This code is contributed by Arnab Kundu |
C#
// C# implementation of the approachusing System;class GFG { // Structure of a linked list node public class Node { public int data; public Node next; }; // Function to common nodes which have // same value node(s) both list static int countCommonNodes(Node head1, Node head2) { // list A Node current1 = head1; // list B Node current2 = head2; // set count = 0 int count = 0; // traverse list A till the end of list while (current1 != null) { // traverse list B till the end of list while (current2 != null) { // if data is match then count increase if (current1.data == current2.data) count++; // increase current pointer for next node current2 = current2.next; } // increase current pointer of list A current1 = current1.next; // initialize list B starting point current2 = head2; } // return count return count; } // Utility function to insert a node at the beginning / static Node push(Node head_ref, int new_data) { Node new_node = new Node(); new_node.data = new_data; new_node.next = head_ref; head_ref = new_node; return head_ref; } // Utility function to print a linked list static void printList(Node head) { while (head != null) { Console.Write(head.data + " "); head = head.next; } Console.WriteLine(); } // Driver code public static void Main(String[] args) { Node head1 = null; Node head2 = null; // Create following linked list // List A = 3 . 4 . 12 . 10 . 17 head1 = push(head1, 17); head1 = push(head1, 10); head1 = push(head1, 12); head1 = push(head1, 4); head1 = push(head1, 3); // List B = 10 . 4 . 8 . 575 . 34 . 12 head2 = push(head2, 12); head2 = push(head2, 34); head2 = push(head2, 575); head2 = push(head2, 8); head2 = push(head2, 4); head2 = push(head2, 10); // print list A Console.Write("Given Linked List A: \n"); printList(head1); // print list B Console.Write("Given Linked List B: \n"); printList(head2); // call function for count common node int count = countCommonNodes(head1, head2); // print number of common node in both list Console.Write("Number of common node in both list is = " + count); }}/* This code contributed by PrinciRaj1992 */ |
Javascript
<script>// Javascript implementation of above approach // Structure of a linked list nodeclass Node{ constructor(val) { this.data = val; this.next = null; }}// Function to common nodes which have// same value node(s) both listfunction countCommonNodes(head1, head2){ // List A var current1 = head1; // List B var current2 = head2; // Set count = 0 var count = 0; // Traverse list A till the end of list while (current1 != null) { // Traverse list B till the end of list while (current2 != null) { // If data is match then count increase if (current1.data == current2.data) count++; // Increase current pointer for next node current2 = current2.next; } // Increase current pointer of list A current1 = current1.next; // Initialize list B starting point current2 = head2; } // Return count return count;}// Utility function to insert a node at the beginningfunction push(head_ref, new_data){ var new_node = new Node(); new_node.data = new_data; new_node.next = head_ref; head_ref = new_node; return head_ref;}// Utility function to prvar a linked listfunction printList(head){ while (head != null) { document.write(head.data + " "); head = head.next; } document.write("<br/>");}// Driver codevar head1 = null;var head2 = null;// Create following linked list// List A = 3 . 4 . 12 . 10 . 17head1 = push(head1, 17);head1 = push(head1, 10);head1 = push(head1, 12);head1 = push(head1, 4);head1 = push(head1, 3);// List B = 10 . 4 . 8 . 575 . 34 . 12head2 = push(head2, 12);head2 = push(head2, 34);head2 = push(head2, 575);head2 = push(head2, 8);head2 = push(head2, 4);head2 = push(head2, 10);// Print list Adocument.write("Given Linked List A: <br/>");printList(head1);// Print list Bdocument.write("Given Linked List B: <br/>");printList(head2);// Call function for count common nodevar count = countCommonNodes(head1, head2);// Print number of common node in both listdocument.write("Number of common node in both list is = " + count);// This code is contributed by gauravrajput1</script> |
Output:
Given Linked List A: 3 4 12 10 17 Given Linked List B: 10 4 8 575 34 12 Number of common node in both list is = 3
Time Complexity: O(M*N), where M is length of list A and N is length of list B
Efficient Solution: Insert all the nodes of linked list A in the unordered_set and then check for each node of linked list B in unordered_set. If found increment the count and return the count at the end.
Below is the implementation of the above approach:
C++
// C++ implementation of above approach#include <bits/stdc++.h>using namespace std;// Structure of a linked list nodestruct Node { int data; struct Node* next;};// Function to common nodes which have// same value node(s) both listint countCommonNodes(struct Node* head1, struct Node* head2){ // list A struct Node* current1 = head1; // list B struct Node* current2 = head2; // set count = 0 int count = 0; // create unordered_set unordered_set<int> map; // traverse list A till the end of list while (current1 != NULL) { // insert list data in map map.insert(current1->data); // increase current pointer of list A current1 = current1->next; } while (current2 != NULL) { // traverse list B till the end of list // if data match then increase count if (map.find(current2->data) != map.end()) count++; // increase current pointer of list B current2 = current2->next; } // return count return count;}/* Utility function to insert a node at the beginning */void push(struct Node** head_ref, int new_data){ struct Node* new_node = new Node; new_node->data = new_data; new_node->next = *head_ref; *head_ref = new_node;}/* Utility function to print a linked list */void printList(struct Node* head){ while (head != NULL) { cout << head->data << " "; head = head->next; } cout << endl;}/* Driver program to test above functions */int main(){ struct Node* head1 = NULL; struct Node* head2 = NULL; /* Create following linked list List A = 3 -> 4 -> 12 -> 10 -> 17 */ push(&head1, 17); push(&head1, 10); push(&head1, 12); push(&head1, 4); push(&head1, 3); // List B = 10 -> 4 -> 8 -> 575 -> 34 -> 12 push(&head2, 12); push(&head2, 34); push(&head2, 575); push(&head2, 8); push(&head2, 4); push(&head2, 10); // print list A cout << "Given Linked List A: \n"; printList(head1); // print list B cout << "Given Linked List B: \n"; printList(head2); // call function for count common node int count = countCommonNodes(head1, head2); // print number of common node in both list cout << "Number of common node in both list is = " << count; return 0;} |
Java
// Java implementation of above approachimport java.util.*;class solution { // static class of a linked list node static class Node { int data; Node next; } // Function to common nodes which have // same value node(s) both list static int countCommonNodes(Node head1, Node head2) { // list A Node current1 = head1; // list B Node current2 = head2; // set count = 0 int count = 0; // create vector Vector<Integer> map = new Vector<Integer>(); // traverse list A till the end of list while (current1 != null) { // insert list data in map map.add(current1.data); // increase current pointer of list A current1 = current1.next; } while (current2 != null) { // traverse list B till the end of list // if data match then increase count if (map.contains(current2.data)) count++; // increase current pointer of list B current2 = current2.next; } // return count return count; } /* Utility function to insert a node at the beginning */ static Node push(Node head_ref, int new_data) { Node new_node = new Node(); new_node.data = new_data; new_node.next = head_ref; head_ref = new_node; return head_ref; } /* Utility function to print a linked list */ static void printList(Node head) { while (head != null) { System.out.print(head.data + " "); head = head.next; } System.out.println(); } /* Driver program to test above functions */ public static void main(String args[]) { Node head1 = null; Node head2 = null; /* Create following linked list List A = 3 . 4 . 12 . 10 . 17 */ head1 = push(head1, 17); head1 = push(head1, 10); head1 = push(head1, 12); head1 = push(head1, 4); head1 = push(head1, 3); // List B = 10 . 4 . 8 . 575 . 34 . 12 head2 = push(head2, 12); head2 = push(head2, 34); head2 = push(head2, 575); head2 = push(head2, 8); head2 = push(head2, 4); head2 = push(head2, 10); // print list A System.out.print("Given Linked List A: \n"); printList(head1); // print list B System.out.print("Given Linked List B: \n"); printList(head2); // call function for count common node int count = countCommonNodes(head1, head2); // print number of common node in both list System.out.print("Number of common node in both list is = " + count); }}// This code is contributed by Arnab Kundu |
Python3
# Python3 implementation of the approach# Structure of a linked list nodeclass Node: def __init__(self): self.data = 0 self.next = None# Function to common nodes which have# same value node(s) both listdef countCommonNodes(head1, head2): # List A current1 = head1 # List B current2 = head2 # Set count = 0 count = 0 # Create unordered_set map_=set() # Traverse list A till the end of list while (current1 != None) : # Add list data in map_ map_.add(current1.data) # Increase current pointer of list A current1 = current1.next while (current2 != None) : # Traverse list B till the end of list # if data match then increase count if ((current2.data) in map_): count = count + 1 # Increase current pointer of list B current2 = current2.next # Return count return count# Utility function to add a node at the beginningdef push(head_ref,new_data): new_node = Node() new_node.data = new_data new_node.next = head_ref head_ref = new_node return head_ref# Utility function to print a linked listdef printList(head): while (head != None) : print(head.data, end = " ") head = head.next # Driver program to test above functionshead1 = Nonehead2 = None# Create following linked list# List A = 3 . 4 . 12 . 10 . 17head1 = push(head1, 17)head1 = push(head1, 10)head1 = push(head1, 12)head1 = push(head1, 4)head1 = push(head1, 3)# List B = 10 . 4 . 8 . 575 . 34 . 12head2 = push(head2, 12)head2 = push(head2, 34)head2 = push(head2, 575)head2 = push(head2, 8)head2 = push(head2, 4)head2 = push(head2, 10)# Print list Aprint("Given Linked List A: ")printList(head1)# Print list Bprint( "\nGiven Linked List B: ")printList(head2)# Call function for count common nodecount = countCommonNodes(head1, head2)# Print number of common node in both listprint("\nNumber of common node in both list is = " , count)# This code is contributed by Arnab Kundu |
C#
// C# implementation of above approachusing System;using System.Collections.Generic;class GFG{ // static class of a linked list node public class Node { public int data; public Node next; } // Function to common nodes which have // same value node(s) both list static int countCommonNodes(Node head1, Node head2) { // list A Node current1 = head1; // list B Node current2 = head2; // set count = 0 int count = 0; // create vector List<int> map = new List<int>(); // traverse list A till the end of list while (current1 != null) { // insert list data in map map.Add(current1.data); // increase current pointer of list A current1 = current1.next; } while (current2 != null) { // traverse list B till the end of list // if data match then increase count if (map.Contains(current2.data)) count++; // increase current pointer of list B current2 = current2.next; } // return count return count; } /* Utility function to insert a node at the beginning */ static Node push(Node head_ref, int new_data) { Node new_node = new Node(); new_node.data = new_data; new_node.next = head_ref; head_ref = new_node; return head_ref; } /* Utility function to print a linked list */ static void printList(Node head) { while (head != null) { Console.Write(head.data + " "); head = head.next; } Console.WriteLine(); } /* Driver code */ public static void Main(String []args) { Node head1 = null; Node head2 = null; /* Create following linked list List A = 3 . 4 . 12 . 10 . 17*/ head1 = push(head1, 17); head1 = push(head1, 10); head1 = push(head1, 12); head1 = push(head1, 4); head1 = push(head1, 3); // List B = 10 . 4 . 8 . 575 . 34 . 12 head2 = push(head2, 12); head2 = push(head2, 34); head2 = push(head2, 575); head2 = push(head2, 8); head2 = push(head2, 4); head2 = push(head2, 10); // print list A Console.Write("Given Linked List A: \n"); printList(head1); // print list B Console.Write("Given Linked List B: \n"); printList(head2); // call function for count common node int count = countCommonNodes(head1, head2); // print number of common node in both list Console.Write("Number of common node in both list is = " + count); }}// This code has been contributed by 29AjayKumar |
Javascript
<script> // JavaScript implementation of above approach // static class of a linked list node class Node { constructor() { this.data = 0; this.next = null; } } // Function to common nodes which have // same value node(s) both list function countCommonNodes(head1, head2) { // list A var current1 = head1; // list B var current2 = head2; // set count = 0 var count = 0; // create vector var map = []; // traverse list A till the end of list while (current1 != null) { // insert list data in map map.push(current1.data); // increase current pointer of list A current1 = current1.next; } while (current2 != null) { // traverse list B till the end of list // if data match then increase count if (map.includes(current2.data)) count++; // increase current pointer of list B current2 = current2.next; } // return count return count; } /* Utility function to insert a node at the beginning */ function push(head_ref, new_data) { var new_node = new Node(); new_node.data = new_data; new_node.next = head_ref; head_ref = new_node; return head_ref; } /* Utility function to print a linked list */ function printList(head) { while (head != null) { document.write(head.data + " "); head = head.next; } document.write("<br>"); } /* Driver code */ var head1 = null; var head2 = null; /* Create following linked list List A = 3 . 4 . 12 . 10 . 17*/ head1 = push(head1, 17); head1 = push(head1, 10); head1 = push(head1, 12); head1 = push(head1, 4); head1 = push(head1, 3); // List B = 10 . 4 . 8 . 575 . 34 . 12 head2 = push(head2, 12); head2 = push(head2, 34); head2 = push(head2, 575); head2 = push(head2, 8); head2 = push(head2, 4); head2 = push(head2, 10); // print list A document.write("Given Linked List A: <br>"); printList(head1); // print list B document.write("Given Linked List B: <br>"); printList(head2); // call function for count common node var count = countCommonNodes(head1, head2); // print number of common node in both list document.write("Number of common node in both list is = " + count); // This code is contributed by rdtank. </script> |
Output:
Given Linked List A: 3 4 12 10 17 Given Linked List B: 10 4 8 575 34 12 Number of common node in both list is = 3
Time Complexity: O(N)
Space Complexity: O(N)
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