Copy a linked list with next and arbit pointer

Problem: Given a doubly linked list with one pointer of each node pointing to the next node just like in a singly linked list. The second pointer(arbit pointer) however can point to any node in the list and not just the previous node. Write a program in to create a copy of this list.

Algorithm:
1. The first step is to create and insert the nodes of copy list after the original node.

2. Now change the arbit pointer of copy list like this

    temp->next->arbit = temp->arbit->next;

where temp is the node in original list and temp2 is the node in copy list)

3. Restore the original list and copy list

    temp->next = temp->next->next;
    temp2->next = temp2->next->next;
   

Implementation:

struct list
{
    struct list *next;
    struct list *arbit;
    int val;
};

struct list *createNode(int val)
{
    struct list* newnode = (struct list*)malloc(sizeof(struct list));
    newnode->val =  val;
    newnode->next = NULL;
    newnode->arbit = NULL;
    return newnode;
}

struct list *copyList(struct list **head)
{
    if(*head==NULL)
    return NULL;
    struct list *copy_list,*temp,*temp2;
    temp = *head;
    //Step 1
    while(temp!=NULL)
    {
         struct list* newnode = createNode(temp->val);
         temp2 = temp->next;
         temp->next = newnode;
         newnode->next = temp2;
         temp = temp->next->next;
    }

    temp = *head;

    //Step 2
    while(temp != NULL && temp->next != NULL)
    {
        temp->next->arbit = temp->arbit->next;
        temp = temp->next->next;
    }

    //Step 3
    temp = *head;
    copy_list = temp->next;
    while(temp != NULL)
    {
        temp2 = temp->next;
        if(temp->next != NULL)
            temp->next = temp->next->next;
        if(temp2->next != NULL)
            temp2->next = temp2->next->next;
        temp = temp->next;
    }
    return copy_list;
 }

Time Complexity : O(N) where N is the number of nodes in list.