C++ linked lists using pointers - beginner question - c++

Question

I have been learning C++ for the last couple of months, and after going through an online course I have started doing some challenges with using raw pointers. I have successfully created class that can initiate and iterate through linked list, with lots of struggle I managed to create method that deletes its element(s), but I am struggling to write a method that would remove all the duplicates.
I have a method such as
void linked_list::remove_dups(){
Node *p = first;
Node *g = first;
while(p!=NULL){
int x = 0;
g = new Node;
g = first;
Node *remove = first;
while (g!=NULL){
if(p->data == g->data){
x++;
}
if(x>1){
remove = new Node;
remove = p;
p = p->next;
remove->next = p->next;
delete remove;
x--;
}
g = g->next;
}
p = p->next;
}
}
where Node is standard
struct Node{
int data;
struct Node *next;
};
first is private member
Node *first;
Can anyone point out what is wrong in this function and explain please? I initiated list with pointers that is
int a[] = {3,5,5,7,65,5,65,65,4,4,15};
and it removed most of the duplicates, but not all. If I add more though, I have EXC_BAD_ACCESS error.
Really appreciate any help!

Answer 1

The main issue is:
p = p->next;
remove->next = p->next;
delete remove;
You try to set remove->next to point to the new next element. But remove is going to be deleted. So the element before remove will still point to something deleted. So, after the delete, you'll have a broken list that contains pointer to deleted elements. Next loop iteration, it will break.
Below is the code I edited as I went. Please note that
remove = new Node;
remove = p;
is not only useless, it leaks memory. You can directly do remove = p;
Just draft code as I went. Not fixed. Proper would be keeping a prev pointer or something.
void linked_list::remove_dups()
{
Node *p = first;
Node *g = first;
while(p!=NULL)
{
// int x = 0; remove this
// g = new Node; remove this
g = first;
Node *remove = first;
while (g != NULL)
{
if(p->data == g->data) // test directly, no x
{
// remove = new Node; remove this
remove = p;
p = p->next;
// remove->next = p->next; You are going to delete this, no point modifying it
delete remove;
}
g = g->next;
}
p = p->next;
}
}

Answer 2

Something I wrote for my custom doubly-linked list that will remove duplicates. Not necessarily the most efficient but it gets the job done.
void unique()
{
std::map<int, int> hash;
Node<T>* temp = this->head;
while(temp)
{
hash[temp->data]++;
if(hash[temp->data] > 1)
{
Node<T>* next = temp->next;
Node<T>* prev = temp->prev;
delete temp;
temp = nullptr;
if(!next)
{
temp = prev;
temp->next = nullptr;
hash[temp->data]--;
break;
}
next->prev = prev;
prev->next = next;
temp = next;
hash[temp->data]--;
}
else
{
temp = temp->next;
}
}
}