Find the common nodes in two singly linked list in C++

Suppose we have two singly-linked lists. We have to find the total number of common nodes in both the singly linked list. So if two lists are like [15, 16, 10, 9, 7, 17], and [15, 16, 40, 6, 9], there are three common nodes.

Traverse both lists up to end of the list using two nested loops, for every node in the list, check if it is matched with any node of the second list or not. If a match is found, then increase the counter, and finally return the count.

Example

Live Demo

#include<iostream>
using namespace std;
class Node {
   public:
      int data;
   Node *next;
};
void prepend(Node** start, int new_data) {
   Node* new_node = new Node;
   new_node->data = new_data;
   new_node->next = NULL;
   if ((*start) != NULL){
      new_node->next = (*start);
      *start = new_node;
   }
   (*start) = new_node;
}
int countCommonNodes(Node** start1, Node** start2) {
   Node* ptr = *start1;
   Node* ptr1 = *start2;
   int count = 0;
   while (ptr != NULL) {
      while (ptr1 != NULL) {
         if (ptr->data == ptr1->data) {
            count++;
            break;
         }
         ptr1 = ptr1->next;
      }
      ptr1 = *start2;
      ptr = ptr->next;
   }
   return count;
}
int main() {
   Node* first = NULL;
   Node* second = NULL;
   prepend(&first, 15);
   prepend(&first, 16);
   prepend(&first, 10);
   prepend(&first, 9);
   prepend(&first, 7);
   prepend(&first, 17);
   prepend(&second, 15);
   prepend(&second, 16);
   prepend(&second, 40);
   prepend(&second, 6);
   prepend(&second, 9);
   cout << "Number of common nodes:" << countCommonNodes(&first, &second);
}