Re: Swap nodes in a linked list

> > However having the index, doesn't
> > that mean that you can simply move to the previous node using that index,
> and
> > then place the removed node in front of that? (so first determine previous
> > node (ElementAt(n-1)) then, addBefore(previousNode)
>
> If I understand it correctly, Enumerable.ElementAt in conjunction with
> a LinkedList will not give you the LinkedListNode, but the actual
> element of the list. Therefore, you can't easily get to the previous
> node from an index using Enumerable.ElementAt.
>
> Using LinkedList.Find to get the node with the element retrieved via
> Enumerable.ElementAt is quite inefficient, because you enumerate the
> list twice; also, it won't work correctly if the list contains
> duplicate values. Therefore, I'd suggest looping over the list _nodes_
> (not its elements) yourself until the node at the specified position
> is found, then you automatically have all the possibilities. (It's not
> as short, but efficient. There doesn't seem to be a simple way to get
> an enumerator for the nodes in a LinkedList, or at least I can't see
> one.)

        Ah, that makes sense. I couldn't reach my .NET 3.5 docs as they are on
another box which wasn't booted up, so I assumed ElementAt() returned the
linkedlistnode.

                FB

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