Multiple Linked Lists w/HeadNode

Here is what I want to do:
1. Create a linked list
2. Create a linked list from those deleted in (1)
3. Create a linked list from those deleted in (2)
Total of 3 singly-linked lists.

1-3 have a HeadNode. I have #1 finished. To create a linked list from those deleted in the first linked list, should I use the same HeadNode - use the same structure? Or is it better to create a new HeadNode for linked list #2? If I create a new HeadNode, then I would have to create a new NodePointer.

I know I'll probably have to alter the generic Delete() node to get those that are deleted to 'go into' the second linked list. - I'm not sure how to do this yet. :confused:

Any ideas on how to approach this? Or if I'm thinking the wrong way? :o

Any help greatly appreciated! Little code snippets would be helpful too! :cool:

Post your current code. (Especially Delete().)

Quote:

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Post your current code. (Especially Delete().)

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That would be too much code. It's your regular generic code of a singly-linked list.

Don't have a 'delete' function. Have a 'remove' function, that removes a node from a list and returns a pointer to it. That way you can do:

Code:

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ptr = remove( somelist, somevalue );
add( someotherlist, ptr );

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Really though, if your 'delete' function is "too much code", then you're doing something really really wrong.


Quzah.

Quote:

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Originally Posted by quzah

Don't have a 'delete' function. Have a 'remove' function, that removes a node from a list and returns a pointer to it. That way you can do:

Code:

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ptr = remove( somelist, somevalue );
add( someotherlist, ptr );

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Really though, if your 'delete' function is "too much code", then you're doing something really really wrong.


Quzah.

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It's generally not a good idea to name your functions the same as ones defined by ANSI C. :p

I like to keep things interesting. ;)


Quzah.

well, normally if u want to deleten a node what will u do

1. search for the particular node which u want to delete
2. on node found u will make the link changes like the following node address is given prev node next.
3. and u will free the node which u want to delete

for u'r case dont free that node try to use that node to create one more linked list .

u will have to use one more headpointer but use the same structure. if u use the same headpointer u will lose old list

hopefully u might have got an idea

ssharish2005

Code:

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ptr = remove( somelist, somevalue );
add( someotherlist, ptr1 );

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I'm guessing your add is same as an

Code:

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insert(someotherlist, ptr1);

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?

I was thinking along the terms of ssharish2005's reply. What do the rest of you think?

Dammit.. I need 3 headpointers... la la la

Well you do need three head pointers if you're creating three seperate lists. So I'm not really sure why you're having problems.

Code:

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struct node *list1, *list2, *list3, *temp;
... do fun stuff ...

temp = extractnode( &list1, somevalue ) /* Pull some node from list1. */
if( temp )
{
    addnode( &list2, temp ); /* Put this node on/in list2. */
}
temp = extractnode( &list2, someothervalue ); /* Pull some node from list2. */
if( temp )
{
    addnode( &list3, temp ); /* Put this node on/in list3. */
}

freelist( list1 );
freelist( list2 );
freelist( list 3 );

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Now all you have to do is write the appropriate functions.


Quzah.

Quote:

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Well you do need three head pointers if you're creating three seperate lists.

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Since I need 3 head pointers, I need 3 nodes with data & next... hmm.. I get it. Because I'm going to loose everything if I use the same nodes & headnodes.. dammit! Thanks for the mini code quzah.

must get back to work and my Viennese Chocolate Cafe..... will be back later if i get stuck.

You're not going to lose anything. Simply remove one node from a list, and return a pointer to it. All you're doing it moving pointers around so nothing points to it. Here, let me illustrate:

Code:

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node1  <-- top of list 1
  |
node2  <-- Node we want to remove
  |
node3  <-- top of the rest of this list
  |
 ...

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Given the above, this is how you remove node2, and keep the rest of the list:

Code:

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temp = node2; <-- get something to keep track of the node we're removing
node1->next = temp->next; <-- make the node before it, point to the node after it
temp->next = NULL; <-- fix the node we're removing so it doesn't point to anything
return temp; <-- return it so we can still use it

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This now gives you the node you've just removed. You haven't deleted it, it's just not in the list. Do this:

Put three cards or pieces of paper in a stack on your desk. Remove the middle piece and hold it in your hand. There. You have done what I have just illustrated. The next piece from the top piece is now the "third" piece. You still have a handle on the middle piece, which you've removed, so you can still do something with it. Such as, putting it on another pile.

Now just pretend you have three seperate piles of paper...

Quzah.

Code:

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temp = node2; <-- get something to keep track of the node we're removing
node1->next = temp->next; <-- make the node before it, point to the node after it
temp->next = NULL; <-- fix the node we're removing so it doesn't point to anything
return temp; <-- return it so we can still use it

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How can I get list #2 to point to temp?

I showed you how two posts prior to this.


Quzah.