C ++循环链表：remove元素

Question

I am done with insertion, search in circular linked list but for removal I am getting compiler errors...

Following is my structure for nodes.

 struct node
 {
     int               p_data;
     struct node*   p_next;

     node(node* head, int data)
     {
           p_next = head;
           p_data = data;
     }

     explicit node(int data)
      {
           p_next = nullptr;
           p_data = data;
      }
 };




 node* remove_circular(node* head, node* target)
 {
      if (head == target->p_next)
      {
           delete head;
           return nullptr;
      }

      auto next_pointer = target->p_next;
      target->p_data = next_pointer->p_data;
      target->p_next = next_pointer->p_next;

      delete target->p_next;
      return target;
 }

and in main function I call

 head = remove_circular(head, head);
 head = remove_circular(head, temp);

this is to remove head element and another element that temp points to. But I am getting errors

Anybody has any idea to remove one element from circular list??

I changed it to delete target->p_next; but now it deletes everything in the list. Any idea???

Answer 1

This is how a circular linked list works:

---

---

Each node points to the next in line, and the tail of the list points to the header node. That's the difference from a circular linked list to a regular linked list (which, in the case above, would make 37 point to a terminator null ).

In the case of your list having only one object, then it should look something like this:

---

---

So, as you can see, there is no object pointing to null anywhere, yet it happens on your code with your explicit constructor (which will run if I write node n = node(12) ).

I suggest you take a look at this link to have a better understanding of how your algorithm should look like.

Answer 2

Once you resolve your compiler error, you are still going to have algorithmic issues. I suggest you draw a circular list on paper and think about the steps required to remove an element. Consider all the cases, for example: empty list, list of 1 item, element not in the list, etc.

Answer 3

You need to consider several things.

1.) the case of an empty list

  if(head == nullptr){//Empty list case
      return nullptr;
  }

2.) The target to be removed is the head node and this is the only node in the list.

  if (head == target && target->p_next == head){
       create a temp node with the data value of target
       target = nullptr;//Since nothing points to target now it is for all intents and purposes deleted from the list but the data is still there so you can do something with it. I assume this is necessary because you return a node *.
       return the temp node
  }

3.) Create a loop that iterates through the entire list. You have something that would only delete the next node which works if you have a two item list and target was the second item.

  auto next_pointer = head->p_next;//could not be target->p_next as this assumed 
  while (next_pointer->p_next != target){//This while loop traverses the list rather than just deleting the next entry.

4.)Inside you loop add a check to see if the list has been traversed and target was never found.

   if (next_pointer->p_next == head){
      return nullptr;
   }//end IF

5.) Inside the loop add the else case which means target was in an arbitrary location in the list. Since I gave you the rest I'll leave you to get this part. It's not hard just a few lines longer than the statements above.