Swapping adjacent elements of linked list - c++

The below is my code to recursive swap the adjacent elements of a linked list. I am losing the pointer to every second element after the swap.
The input is 1->2->3->4->5->6->7, I expected the output 2->1->4->3->6->5->7,
but my output is 1->3->5->7.
void nodelist::swap(node* head)
{
node* temp = head->next;
if (head->next!= nullptr)
{
node* temp2 = temp->next;
temp->next = head;
head->next = temp2;
head = head->next;
temp = nullptr;
temp2 = nullptr;
swap(head);
}
}
Any help would be appreciated,thanks in advance.

In fact it is enough to swap only the data members of nodes. There is no need to swap the pointers themselves.
Nevertheless if to use your approach then the function can look like
void SwapList( node *head )
{
if ( head != nullptr && head->next != nullptr )
{
node *next = head->next;
std::swap( *head, *next );
std::swap( head->next, next->next );
SwapList( head->next->next );
}
}
Here is a demonstrative program
#include <iostream>
#include <utility>
struct node
{
int value;
node *next;
};
node * AddNode( node *head, int value )
{
head = new node { value, head };
return head;
}
void PrintList( node *head )
{
for ( ; head != nullptr; head = head->next )
{
std::cout << head->value << ' ';
}
}
void SwapList( node *head )
{
if ( head != nullptr && head->next != nullptr )
{
node *next = head->next;
std::swap( *head, *next );
std::swap( head->next, next->next );
SwapList( head->next->next );
}
}
int main()
{
node *head = nullptr;
for ( int i = 10; i != 0; )
{
head = AddNode( head, --i );
}
PrintList( head );
std::cout << std::endl;
SwapList( head );
PrintList( head );
std::cout << std::endl;
return 0;
}
The output is
0 1 2 3 4 5 6 7 8 9
1 0 3 2 5 4 7 6 9 8
You can use the shown function as a template (or base) for your function.

With no recursion:
void swap(node **head)
{
while (*head && (*head)->next)
{
node* tmp = *head;
*head = tmp->next;
tmp->next = (*head)->next;
(*head)->next = tmp;
head = &tmp->next;
}
}
Invoke swap( & list_head_ptr).
Alternatively, you can pass the head pointer by reference-to-pointer and utilize a local pointer-to-pointer member:
void swap(node*& head)
{
node **pp = &head;
while (*pp && (*pp)->next)
{
node* tmp = *pp;
*pp = tmp->next;
tmp->next = (*pp)->next;
(*pp)->next = tmp;
pp = &tmp->next;
}
}
and invoke as swap(list_head_ptr). Either method works.

Using recursion:
void nodelist::swap(node** head) {
if (!*head || !(*head)->next) return;
node* const sw = (*head)->next;
(*head)->next = sw->next;
sw->next = *head;
*head = sw;
swap(&(sw->next->next));
}

If head is the pointer which stores the address of firstNode (value=1), then try following function:
void nodelist::swap(node* head){
node* temp = head->next; //head->next is first-node which needs to switch with it's next node
if (temp!= nullptr && temp->next!=nullptr){
head->next=temp->next; //move second node to first
temp->next = head->next->next; //put second's next in first's
head->next->next = temp; //and first will be second's next
temp = nullptr; // swaping done
swap(head->next->next); //do it for next couple
}
}
http://coliru.stacked-crooked.com/a/e1cc0d02b5599da4
OR
http://coliru.stacked-crooked.com/a/a1e200b687825d80
If head itself is the firstNode (value=1), then passing head by value will not work, either you need to pass it by address/reference OR do it like in following link:
http://coliru.stacked-crooked.com/a/a1e200b687825d80

Related

Reverse a linked list using recursion

The void reve(struct Node *head) and display(struct Node *head) methods take one argument - the head of the linked list. I want to print the whole linked list but my display function print only 4.
#include <iostream>
using namespace std;
struct Node {
int data;
struct Node *link;
};
void display(struct Node *head) {
if (head == NULL) {
return;
}
cout << head->data << "\t";
display(head->link);
//head = head->link;
}
struct Node *reve(struct Node *head) {
struct Node *p = head;
if (p->link == NULL) {
head = p;
return head;
}
reve(p->link);
struct Node *temp = p->link;
temp->link = p;
p->link = NULL;
}
struct Node *insert(struct Node *head, int new_data) {
Node *new_node = new Node();
new_node->data = new_data;
new_node->link = head;
head = new_node;
return head;
}
int main() {
Node *head = NULL;
head = insert(head, 1);
head = insert(head, 2);
head = insert(head, 3);
head = insert(head, 4);
cout << "The linked list is: ";
//display(head);
head = reve(head);
display(head);
return 0;
}
Output
If you want the recursive way:
Node* reverse(Node* head)
{
if (head == NULL || head->next == NULL)
return head;
/* reverse the rest list and put
the first element at the end */
Node* rest = reverse(head->next);
head->next->next = head;
head->next = NULL;
/* fix the head pointer */
return rest;
}
/* Function to print linked list */
void print()
{
struct Node* temp = head;
while (temp != NULL) {
cout << temp->data << " ";
temp = temp->next;
}
}
The function reve does not return a value if p->link is not NULL.
Since head has more than 1 element, head = reve(head); has undefined behavior.
Reversing a linked list is much easier to implemented in a simple loop than with recursion:
struct Node *reve(struct Node *p) {
if (p != NULL) {
struct Node *prev = NULL;
while (p->link) {
struct Node *next = p->link;
p->link = prev;
prev = p;
p = next;
}
}
return p;
}
If your task requires recursion, you can make a extract the first node, reverse the remainder of the list and append the first node. Beware that this is not tail recursion, hence any sufficiently long list may cause a stack overflow.
struct Node *reve(struct Node *head) {
if (head != NULL && head->link != NULL) {
struct Node *first = head;
struct Node *second = head->link;
head = reve(second);
first->link = NULL; // unlink the first node
second->link = first; // append the first node
}
return head;
}
In C++ you need not to use keywords struct or class when an already declared structure or a class is used as a type specifier.
The function reve has undefined behavior.
First of all head can be equal to nullptr. In this case this statement
if (p->link == NULL) {
invokes undefined behavior.
Secondly the function returns nothing in the case when p->link is not equal to nullptr.
//...
reve(p->link);
struct Node *temp = p->link;
temp->link = p;
p->link = NULL;
}
Here is a demonstrative program that shows how the functions can be implemented. I used your C approach of including keyword struct when the structure is used as a type specifier.
#include <iostream>
struct Node
{
int data;
struct Node *link;
};
struct Node * insert( struct Node *head, int data )
{
return head = new Node{ data, head };
}
struct Node * reverse( struct Node *head )
{
if ( head && head->link )
{
struct Node *tail = head;
head = reverse( head->link );
tail->link->link = tail;
tail->link = nullptr;
}
return head;
}
std::ostream & display( struct Node *head, std::ostream &os = std::cout )
{
if ( head )
{
os << head->data;
if ( head->link )
{
os << '\t';
display( head->link, os );
}
}
return os;
}
int main()
{
struct Node *head = nullptr;
const int N = 10;
for ( int i = 0; i < N; i++ )
{
head = insert( head, i );
}
display( head ) << '\n';
head = reverse( head );
display( head ) << '\n';
return 0;
}
The program output is
9 8 7 6 5 4 3 2 1 0
0 1 2 3 4 5 6 7 8 9
display is fine.
First thing I have notices is that you are trying to modify a copied value. For example, line 16. This code has no effect.
Note that you have a bug on insert: You return head instead of new_node.
Your version fails for lists with more than 1 item. reve() is supposed to return the last node of the original list, which you do not, hence lastNode would not point to the last node of the reversed list. So my advice is that you keep it aside.
So, reve:
struct Node* reve(struct Node* head) {
if (head->link == NULL) {
return head;
}
struct Node* lastNode = reve(head->link);
lastNode->link = head;
head->link = NULL;
return head;
}
and main:
int main() {
Node* head = NULL;
Node* last_node = head = insert(head, 1);
head = insert(head, 2);
head = insert(head, 3);
head = insert(head, 4);
head = reve(head);
cout << "The linked list is: ";
// Now, last_node is the new head
display(last_node);
return 0;
}

Error in creating a linked list from another linked list?

I'm creating a linked list from another linked list. But second linked list is not getting formed and there's a memory leak message on running the program.
Here's a section of the code that's troubling-
Node *rearrangeLinkedList(Node *head){
printLinkedList(head);
int lengthoflist = 0;
Node *temp = head;
while (temp!=NULL)
{
lengthoflist++;
temp=temp->next;
}
Node *secondList = NULL;
// just a variable node to store the head of second linked list-
Node *headOfSecondList = secondList;
int count=1;
while (count<=lengthoflist/2)
{
Node *temproary = new Node();
temproary->data=head->data;
temproary->next=NULL;
secondList=temproary;
secondList=secondList->next;
head=head->next;
count++;
}
printLinkedList(headOfSecondList);
}
printLinkedList() function is perfectly printing out the incoming list but not the second linked list.
After
Node *secondList = NULL;
Node *headOfSecondList = secondList;
you don't modify headOfSecondList any more. It will still be NULL when you call
printLinkedList(headOfSecondList); // => printLinkedList(NULL);
But you have another error in the copy-function:
while (count<=lengthoflist / 2)
{
Node *temproary = new Node();
temproary->data=head->data;
temproary->next=NULL;
secondList=temproary; // assign secondList
secondList=secondList->next; // secondList->next is temporary->next is NULL!!
head=head->next;
count++;
}
Here you create a bunch of nodes that all have a next of NULL. You do indeed leak memory here. secondList gets set to NULL at the end of each iteration and when temporary goes out of scope you don't have any pointers to the allocated memory left.
The following should work
// Build first node
Node *secondList = new Node();
secondList->data = head->data;
// advance by one
head = head->next;
// Now this points to the real head instead of NULL
Node *headOfSecondList = secondList;
int count=1;
while (count<=lengthoflist / 2 - 1 ) // -1 since we already handled the head above
{
Node *temproary = new Node(); // new node
temproary->data = head->data; // set data
temproary->next = NULL; // we have no next yet
secondList->next = temproary; // append temporary to secondList
secondList = secondList->next; //advance secondList
head = head->next; // advance head
count++;
}
printLinkedList(headOfSecondList);
I have skipped some validation here, but I hope the basic concept is clearer now.
If I have understood correctly the function tries to build a new list from the first half of nodes of an existed list.
If so then there is no need to calculate the number of nodes in the source list. This is inefficient.
You declared the function having the return type Node *.
Node *rearrangeLinkedList(Node *head );
but the function returns nothing.
Within the function the variable headOfSecondList is set once to nullptr and is never changed.
Node *secondList = NULL;
Node *headOfSecondList = secondList;
Within the while loop new nodes are not chained in a list. There is always changed the variable secondList and its data member next is always set to NULL. So there are numerous memory leaks.
while (count<=lengthoflist/2)
{
Node *temproary = new Node();
temproary->data=head->data;
temproary->next=NULL;
secondList=temproary;
secondList=secondList->next;
head=head->next;
count++;
}
The function can be written the following way.
Node * rearrangeLinkedList( Node *head )
{
Node *new_head = nullptr;
Node **tail = &new_head;
Node *first = head, *current = head;
while ( current != nullptr && ( current = current->next ) != nullptr )
{
current = current->next;
*tail = new Node();
( *tail )->data = first->data;
( *tail )->next = nullptr;
first = first->next;
tail = &( *tail )->next;
}
return new_head;
}
To demonstrate the approach without counting the number of nodes in the source list that as I already pointed out is inefficient here is a demonstrative program with a class template List.
#include <iostream>
template <typename T>
class List
{
private:
struct Node
{
T data;
Node *next;
} *head = nullptr;
public:
List() = default;
~List()
{
while ( head )
{
Node *current = head;
head = head->next;
delete current;
}
}
List( const List<T> & ) = delete;
List<T> & operator =( const List<T> & ) = delete;
void push_front( const T &data )
{
head = new Node { data, head };
}
List<T> & extract_half( List<T> &list ) const
{
Node **tail = &list.head;
while ( *tail ) tail = &( *tail )->next;
Node *first = this->head, *current = this->head;
while ( current != nullptr && ( current = current->next ) != nullptr )
{
current = current->next;
*tail = new Node { first->data, nullptr };
first = first->next;
tail = &( *tail )->next;
}
return list;
}
friend std::ostream & operator <<( std::ostream &os, const List &list )
{
for ( Node *current = list.head; current; current = current->next )
{
os << current->data << " -> ";
}
return os << "null";
}
};
int main()
{
List<int> list1;
const int N = 10;
for ( int i = N; i != 0; )
{
list1.push_front( --i );
}
std::cout << list1 << '\n';
List<int> list2;
list1.extract_half( list2 );
std::cout << list1 << '\n';
std::cout << list2 << '\n';
return 0;
}
The program output is
0 -> 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> null
0 -> 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> null
0 -> 1 -> 2 -> 3 -> 4 -> null

Singly Linked List Infinite Loop

It's been a week since i started learning about linked list and i only managed to learn about singly linked list. So today i implemented the linked list which i learned in c++ and while i tried to run it the code goes into an infinite loop of some random numbers. I tried debugging the code but i coudn't find whats so ever is wrong with the code. The code is below. Help is appreciated.Thanks
#include <iostream>
using namespace std;
struct node{
int data;
node * next;
};
class singly{
private:
node * head,*tail;
public:
singly(){
head=NULL;
tail=NULL;
}
void createNode(int value){
node * temp = new node;
temp->data=value;
temp->next=NULL;
if(head==NULL){
head=temp;
tail=temp;
temp=NULL;
}
else{
tail->next=temp;
tail=temp;
}
}
void display(){
node * temp = new node;
head=temp;
while(temp!=NULL){
cout << temp->data << "\t" << endl;
temp->next=temp;
}
}
void insert_end(int value){
node*newnode = new node;
node*temp = new node;
newnode->data=value;
newnode->next=NULL;
temp=head;
while(temp->next!=NULL){
temp = temp->next;
}
temp->next=newnode;
}
void delete_node(){
node*current = new node;
node*previous = new node;
current = head;
while(current->next!=NULL){
previous=current;
current=current->next;
}
tail=previous;
previous->next=NULL;
delete current;
}
};
int main(){
singly lists;
lists.createNode(32);
lists.createNode(654);
lists.createNode(34);
lists.createNode(234);
cout<<"\n--------------------------------------------------\n";
cout<<"---------------Displaying All nodes---------------";
cout<<"\n--------------------------------------------------\n";
lists.display();
cout<<"\n--------------------------------------------------\n";
cout<<"-----------------Inserting At End-----------------";
cout<<"\n--------------------------------------------------\n";
lists.createNode(55);
lists.display();
cout<<"\n--------------------------------------------------\n";
cout<<"-----------------Deleing At End-------------------";
cout<<"\n--------------------------------------------------\n";
lists.delete_node();
lists.display();
}
The member function display does not make sense.
It overwtites the data member head with uninitialized newly created temp.
node * temp = new node;
head=temp;
so the function invokes undefined behavior.
The function can look like
void display()
{
for ( node * temp = head; temp != nullptr; temp = temp->next )
{
cout << temp->data << "\t";
}
}
Or it is better to define it the following way
std::ostream & display( std::ostream &os = std::cout )
{
for ( node * temp = head; temp != nullptr; temp = temp->next )
{
os << temp->data << "\t";
}
return os;
}
The data member insert_end is also wrong. It does not take into account that head and tail can be equalto nullptr and does not change them.
The function can be defined the following way
void insert_end(int value)
{
node *newnode = new node { value, nullptr };
if ( tail == nullptr )
{
head = tail = newnode;
}
else
{
tail = tail->next = newnode;
}
}
The member function delete_node firstly does not make sense for a singly-linked list and again is wrong and invokes undefined behavior. The function should remove the first node from the list.
Nevertheless if you want to remove the last node from the list then the function can look like
void delete_node()
{
if ( head != nullptr )
{
tail = nullptr;
node *current = head;
while ( current->next )
{
tail = current;
current = current->next;
}
if ( tail == nullptr )
{
head = tail;
}
else
{
tail->next = nullptr;
}
delete current;
}
}
For starters, display() is wrong. You want the update to be temp = temp->next; and it can also be initialized as node * temp = head hence not requiring the second line.
Your delete_node() can be re-written to:
if (head->next == NULL) // handles the case that it consists of 1 element
{
delete head;
head = NULL;
}
else
{
node *nextToEnd = head;
node *end = head->next;
while (end->next != NULL)
{
nextToEnd = end;
end = end->next;
}
delete end;
nextToEnd->next = NULL;
}
As stated in the comments, review the use of the new keyword

Swap Adjacent nodes in a Linked List Only by Manipulating pointers

I am trying to swap the adjacent nodes of a linked list i.e.
1->2->3->4->5 becomes 2->1->4->3->5
My function is:
node * swapper(node * &head)
{
if (head == NULL || head->next == NULL) return head;
node * t = head;
head= head->next;
head->next = t;
t->next = head->next;
node *previous = head->next->next, *current = previous->next;
while (current!=NULL&&previous!=NULL)
{
node * t1 = current,*t2=previous;
current->next = previous;
previous->next = t1->next;
previous = t1->next;
current = previous->next;
}
return head;
}
I know it can be done by swapping values but I have to do it in Constant space and without swapping the values.
I can't find why my function is not working.
The first thing that I can notice is that you need to swap those two lines:
head->next = t;
t->next = head->next;
because you are saying head->next = t so you are losing the connection to the rest of the linked list.
Also, inside the loop. There are several mistakes:
1- You're changing the next of current before obtaining it in previous, which means you're losing the link (like above)
2- You're not connecting them to the nodes that are before them.
My five cents.:)
Here is a demonstrative program that shows how to write a recursive function.
#include <iostream>
#include <functional>
struct node
{
int data;
struct node *next;
} *head;
void push_front( node * &head, int x )
{
head = new node { x, head };
}
void display( node *head )
{
for ( node *current = head; current; current = current->next )
{
std::cout << current->data << ' ';
}
std::cout << std::endl;
}
node * swapper( node * &head )
{
if ( head && head->next )
{
node *tmp = std::exchange(head, head->next );
std::exchange( tmp->next, std::exchange( tmp->next->next, tmp ) );
swapper( head->next->next );
}
return head;
}
int main()
{
const int N = 10;
for ( int i = N; i != 0; ) push_front( head, --i );
display( head );
display( swapper( head ) );
}
The program output is the following
0 1 2 3 4 5 6 7 8 9
1 0 3 2 5 4 7 6 9 8
Take into account that not all compilers support function std::exchange. So you will need to write it yourself.:)
If to write the main the following way
int main()
{
const int N = 10;
for ( int i = N; i != 0; ) push_front( head, --i );
display( head );
for ( node **current = &head; *current && ( *current )->next; current = &( *current )->next )
{
swapper( *current );
}
display( head );
}
then the following interesting output can be obtained.:)
0 1 2 3 4 5 6 7 8 9
1 3 5 7 9 8 6 4 2 0
node * swapper(node * &head)
{
if (head == NULL || head->next == NULL) return head;
node * t = head;
head= head->next;
t->next = head->next;
head->next = t;
node *previous = head->next->next, *current,*parent=head->next;
while (previous&&(current = previous->next))
{
node * t1 = current,*t2=previous;
previous->next = t1->next;
current->next = previous;
parent->next= current;
previous = t2->next;
//current = previous->next;
parent=t2;
}
return head;
}

Merging 2 linked lists and appending to the end of linked lists c++

I don't have much so far but I am trying to get the hang of using linked lists.
Struct:
struct Node
{
int value;
Node *next;
};
How can I add a node to the end of the list? I am just trying to take in a pointer for the head of a list and an int value to add in as a new node. When I try running what I have currently I get an exception.
void addNode(Node* head, int x)
{
Node* temp = new Node;
temp->data = x;
temp->next = NULL;
if(!head)
{
head = temp;
return;
}
else
{
Node* last = head;
while(last->next)
last=last->next;
last->next = temp;
}
}
I haven't really begun to work on merging the two lists. I just know that I need to take in 2 linked lists (or pointers to the head of 2 linked lists?) and then run through the lists for all the nodes.
E.G: Linked list 1 has 3 nodes: 4, 10, 20.
Linked List 2 has 4 nodes: 2, 5, 15, 60.
The merge list function would results in a new linked list with 2,4,5,10,15,20,60 as the nodes.
EDIT: In my main, I am calling the addNode function like so:
Node *head = new Node;
insertAtEnd(head,20);
Is that correct or could that be the cause of the exception?
By doing this:
void addNode(Node* head, int x)
// here ---------^
and then later this:
head = temp; // here
you're simply modifying the local head pointer, which took on the address value passed from the caller. Since head is not an actual reference to a pointer (it's just a pointer), the result is the caller's pointer passed as head remains unaltered. You never append your allocated node to your list, leak memory, it becomes a sad day...
Pass the pointer by reference instead. Fixing that, then fixing the invalid data member, which should actually be value and a pointer-to-pointer for walking the list to find the end, the result could look something like this:
#include <iostream>
struct Node
{
int value;
Node *next;
};
void addNode(Node*& head, int x)
{
Node **pp = &head;
while (*pp)
pp = &(*pp)->next;
*pp = new Node;
(*pp)->value = x;
(*pp)->next = nullptr;
}
void printList(const Node *head)
{
for (; head; head = head->next)
std::cout << head->value << ' ';
std::cout << '\n';
}
void freeList(Node *&head)
{
while (head)
{
Node *p = head;
head = p->next;
delete p;
}
}
int main()
{
Node *head = nullptr;
for (int i=1; i<=5; ++i)
addNode(head, i);
printList(head);
freeList(head);
}
Output
1 2 3 4 5
I leave the task of implementing an actual merge to you, but this should be enough to get you a manageable list up and running.
Update: From the OP's edited question:
Node *head = new Node;
insertAtEnd(head,20);
Apart from now-being a completely different named function, your node is default-initialized. In your case that means the resulting Node from new Node; has indeterminate values for both value and next. You're then passing that to your function, which assumes a determinate value (null) to terminate your loop.
This can be fixed any number of ways; the mechanics of the code above is one such way. There is no need to pre-allocate a head node in the first place if the list management code is of the understanding that NULL means no-list. Your addNode original post seemed to at-least-try to follow that mantra.
Declare the function the following way
void addNode( Node* &head, int x) ;
And instead of this code snippet
Node *head = new Node;
insertAtEnd(head,20);
You have to call the function the first time the following way
Node *head = nullptr; // or NULL
addNode(head,20);
Notice that there is no function with name insertAtEnd in your post. There is function addNode.:)
If you need to merge two lists then you can use this demonstrative program as a sample. Of course you will need to add some other functions as for example deleting lists that to get a complete project.
#include <iostream>
struct Node
{
int value;
Node *next;
};
Node * insert( Node *current, int value )
{
Node *tmp;
if ( current == nullptr )
{
tmp = new Node { value, nullptr };
}
else
{
tmp = new Node { value, current->next };
current->next = tmp;
}
return tmp;
}
std::ostream & display( Node *head,
std::ostream &os = std::cout,
const char *delimiter = " " )
{
for ( ; head; head = head->next ) os << head->value << delimiter;
return os;
}
Node * merge( Node * &head1, Node * &head2 )
{
Node *new_head = nullptr;
Node *current = nullptr;
while ( head1 != nullptr && head2 != nullptr )
{
Node *tmp;
if ( head2->value < head1->value )
{
tmp = head2;
head2 = head2->next;
}
else
{
tmp = head1;
head1 = head1->next;
}
tmp->next = nullptr;
if ( new_head == nullptr )
{
new_head = tmp;
current = new_head;
}
else
{
current->next = tmp;
current = current->next;
}
}
if ( head1 != nullptr ) new_head == nullptr ? new_head : current->next = head1;
if ( head2 != nullptr ) new_head == nullptr ? new_head : current->next = head2;
head2 = nullptr;
head1 = new_head;
return new_head;
}
int main()
{
Node *list1 = nullptr;
Node *list2 = nullptr;
list1 = insert( list1, 4 );
insert( insert( list1, 10 ), 20 );
display( list1, std::cout << "List1: " ) << std::endl;
list2 = insert( list2, 2 );
insert( insert( insert( list2, 5 ), 15 ), 60 );
display( list2, std::cout << "List2: " ) << std::endl;
std::cout << std::endl;
merge( list1, list2 );
display( list1, std::cout << "List1: " ) << std::endl;
display( list2, std::cout << "List2: " ) << std::endl;
return 0;
}
The program output is
List1: 4 10 20
List2: 2 5 15 60
List1: 2 4 5 10 15 20 60
List2:
this may be a cause of exception:
struct Node
{
int value; <----- Node structure has value property
Node *next;
};
Node* temp = new Node;
temp->data = x; <------ Assigning to data property of Node which does not exists
temp->next = NULL;
To add list you may use same approach
void addNode(Node* head, Node* head2)
{
Node* last = head;
while(last->next) last=last->next;
last->next = head2;
}
EDIT: In my main, I am calling the addNode function like so:
Node *head = new Node;
insertAtEnd(head,20);
This is wrong. You didn't initialize head->next, so within insertAtEnd the code while(last->next) last=last->next; will attempt to compare uninitialized pointer and if it isn't null, will dereference it. This will likely crash your program rather than throw an exception though. Then again, it's undefined behaviour, so anything may happen.
Since your insert function already covers the case of inserting to empty list, I would simply call
head = nullptr;
insertAtEnd(head,20)`;
Besides that, there's the bug of never updating the head pointer outside the function, which has already been covered in other answers.