C ++了解一个循环的双向链表

Question

I'm having a problem figuring out how to set my ptr's for prev and next when I add a node with at least one existing node. Adding the first node is easy, just setting the ptrs to the front. I need help viewing this mentally, also this program is a queue so each node gets added to the back of the list.

if (Empty())
    {
        front = new qnode;
        front->next=front;
        front->prev=front;
        front->data = item;
    }
    else if (front->prev=front)
    {
        front->prev = new qnode;
        front->prev->next= front;
        front->next=front->prev;
        front->prev->data = item;
    }
    else
    {

    }

What i have now still not getting it

else
    {
        front->prev= new qnode;
        front->prev->data= item;
        front->prev->next=front;
        front->prev=front->prev->prev;

    }

Answer 1

I hope this image helps a bit

I have created pictures for 1 item 2 items and 3 items

the pointers just point to the actual object meaning the black rectangle is the whole object if front being blue and prev being brown(those are just there as references)

I really hope this helps linked list can get really tricky and drawing always helps me.

so to add the item at the front of the list you have some code like this:

 //ok first I'll define some variables for you
 //last === the last node in the list
 //head === the first node in the list
 //node === the new node you are adding;
 qnode node = new qnode;
 node.data = data; //whatever data you are holding
 node->next = last; //last element in the list since it is circular;
 node->prev = head; //you want the new node to point the the first node since it's getting added before that;
 head->next = node; //you want the head of the node to point to the new node not the last item
 last->prev = node; //last node now should point to the new node you just added not the head;

Answer 2

I am not sure why you need an else if , as I think the case with one element is not different from that with multiple elements. In any case you have a syntax error there - you wanted to write == instead of = .

More to the point, what you want to do is:

• Create a new qnode
• Set the data of the new node to the item
• Set the next of the new node to the front
• Set the prev of the new node to front->prev
• Set the prev of the front to the new node.

You may check for yourself on a piece of paper how this works.

Answer 3

In a circular doubly linked list, you have 2 pointers, one to the head and one to the tail (last element in the list). If a element has a next, the next element's prev should be the element pointing at it through it's next element. Your head pointer should never move unless the head element is removed. Your tail will move along always ensuring that the element it points to has head as it's next element. Hopefully this helps

Answer 4

Since you always add at the end, the case of a queue with a single element is identical with the case of a list with multiple elements. Take a paper and a pen, make some sketches and it will come very easy to you.