How to delete two items in a row in a linked list

Question

void delete_double (LN*& l) {
    if (l == nullptr)
        return;

    LN *p = l;
    while ( p -> next != nullptr && p -> next -&g         


        

Answer 1

I have simplified the code above, the logic you have above also wont help you remove multiple duplicates. So lets look at the code below and dissect it:

   void delete_double(LN*& l) {

        if (l == nullptr)
            return;

        LN *p = l;
        LN dummy(0);
        dummy.next = l;
        p = &dummy;

        LN *temp;
        LN *duplicate;
        LN *prev;

        while (p != nullptr && p->next != nullptr)
        {
            temp = p;
            while (p != nullptr && temp->next != nullptr)
            {
                if (p->value == temp->next->value)
                {
                    duplicate = temp->next;
                    temp->next = temp->next->next;
                    delete duplicate;

                    duplicate = p;
                    prev->next = p->next;
                    p = prev;
                    delete duplicate;

                    temp = p;
                }
                else
                {
                    break;
                }
            }
            prev = p;
            p = p->next;
        }

        l = dummy.next;
    }

There seems to be a need for a dummy node at the start because if we have 1 -> 1 - > 2 we need to delete the first two and point to the correct head which is a 2. In order to avoid this confusion is better to keep a dummy node at the start and at the end just set the output of your list to be p = dummy.next, which is the actual start of your list.

I have defined some temporaries, temp and duplicate, temp to help me navigate further in the list and duplicate to hold the duplicate value, move the pointer to next and delete the node. prev is the previous pointer to the node just before the duplicates.

Each node in the list, temp = p I move forward until looking for the adjacent match p->value == temp->next->value if there is a match I delete the current node and the one I found ahead of it. I use the prev tracker to restore the order of the list by correctly setting its next, else I break from my inner loop and move on to the next value, the outer loop p = p->next.

I am not sure about your LN struct so I have gone ahead with what I think it is.

Demo Link