I am not getting how to concatenate two Linked List - c++

Question

I am not getting how to run the concatenate function using both Node and LinkedList as classes. If anyone knows how to do it, please let me know.
Here I have created two classes one for linked list and the former one for Node creation.
Using create function and passing array as Linked list input. Also, I am getting the headref using getHead() and getHead2() functions which give the starting pointer of first and second Linked List respectively.
#include <iostream>
using namespace std;
class Node{
public:
int data;
Node* next;
Node(){
data=0;
next=NULL;
}
Node(int data){
this->data=data;
this->next=NULL;
}
};
class LL{
Node* first, *second;
public:
LL(){
first=second=NULL;
}
void create(int arr[], int n){
Node* t, *last;
first=new Node();
first->data= arr[0];
first->next=NULL;
last=first;
for(int i=1;i<n;i++){
t=new Node();
t->data=arr[i];
t->next=NULL;
last->next=t;
last=t;
}
}
void create2(int arr[], int n){
Node* t, *last;
second=new Node();
second->data= arr[0];
second->next=NULL;
last=second;
for(int i=1;i<n;i++){
t=new Node();
t->data=arr[i];
t->next=NULL;
last->next=t;
last=t;
}
}
Node* getHead(){
return first;
}
Node* getHead2(){
return second;
}
void display(Node* p){
while(p){
cout<<p->data<<" ";
p=p->next;
}
cout<<endl;
}
void concatLL(){
Node* p=first;
while(p->next){
p=p->next;
}
p->next=second;
second=NULL;
}
};
int main()
{
LL l,l2;
int arr[]={1,2,3,4,5,6,7};
int arr2[]={2,5,7,8,9};
int n=sizeof(arr)/sizeof(arr[0]);
int n2=sizeof(arr2)/sizeof(arr2[0]);
l.create(arr,n);
l2.create2(arr2,n2);
cout<<"Displaying first LL"<<endl;
l.display(l.getHead());
cout<<"Displaying second LL"<<endl;
l2.display(l2.getHead2());
cout<<"Displaying Linked list after concatination"<<endl;
l.concatLL();
l.display(l.getHead());
return 0;
}

Answer 1

p is equal to NULL when executing p->next=second; in concatLL() causing undefined behaviour since the while loop runs until p == NULL. You should use while(p->next) instead of while(p) and check that first is not a null pointer before.

Answer 2

A linked list is still a Node, and as others said, at the end of your while loop p is pointed to NULL. You can't NULL->second=second neither NULL=second, so change your while loop stop condition:
void concatLL(){
Node* p=first;
while(p->next){
p=p->next;
}
p->next=second;
second=NULL;
}
};
Here is my solution:
#include <iostream>
using namespace std;
struct Node {
Node* next;
int value;
Node(int value) {
this->value = value;
this->next = NULL;
}
Node(int n, int a[]) {
value = a[0];
next = NULL;
Node* p = this;
for (int i = 1; i < n; i++) {
p->next = new Node(a[i]);
p = p->next;
}
}
void concat(Node* head) {
Node* p = this;
while (p->next != NULL) {
p = p->next;
}
p->next = head;
}
void display() {
Node* current = this;
while (current != nullptr) {
cout << current->value << " ";
current = current->next;
}
cout << endl;
}
};
int main() {
int a[] = { 1, 2, 3, 4, 5 };
int b[] = { 6, 7, 8, 9, 10 };
Node* list1 = new Node(5, a);
Node* list2 = new Node(5, b);
list1->display();
list2->display();
list1->concat(list2);
list1->display();
return 0;
}