how to copy a linkedlist

Copy link-list, If You Can

Make a seprate copy of a linked list that has an extra pointer than theregular ones. This pointer points to any randam  node in thelist.i.e.::node structure is---
struct node
{
  type element
  node  *next;
  node  *other;
};
You are provided with head pointer
eg. *other pointer of first node can point to middle node and *other pointer of last node can point to frist node

I need not remind u to keep in mind the complexity..

Holy_Sin
Thursday, June 29, 2006

 

Deleting … Approving …

 

Yes, can be done in O(n) time and O(1) space, not counting input and output lists...

Here is come C like code..

// Edge case/error detection missing.
LL * Copy (LL *head){

  LL *tmp = head;
  LL *prev = NULL;
  LL *cur = NULL;

  LL *savedNext = NULL;

  // Interleave the new and old lists.
  while (tmp){

    cur = AllocNewLL();

    savedNext = tmp->next;

    tmp->next = cur;
    cur->next = savedNext;
   
    tmp = savedNext;
  }

  tmp = head;
 
  // Get the correct other pointers for the new list.
  while(tmp){

    cur = tmp->next;
    cur->other = tmp->other->next;
    tmp = cur->next;
  }
 
  LL * copiedList = head->next;

  // Restore the original list.
  while (tmp){

    cur = tmp->next;
    tmp->next = cur->next;
    tmp = cur->next;
  }

  return copiedList;

}

Basically, if original list was A1 -> A2 -> ... -> An

The we make a copy of node Ai. Call that node Bi.

We first make the list look like

A1 -> B1 -> A2 -> B2 -> ... -> An -> Bn (first while loop)

Nowto get the correct other pointer of Bi, we take the other pointer ofAi, and get the next. i.e say the other of Ai is pointing to Aj. Thenwhen we follow the next of Aj we get Bj which should be the other ofBi. (second while loop)

Now we restore the input list back to its form:
A1 -> A2 -> ...-> An. (third while loop)