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 the
regular ones. This pointer points to any randam node in the
list.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)

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

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