Copy link-list, If You Can
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..
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)