Re: java.util.LinkedList.iterator().remove() time complexity

Joshua Cranmer <Pidgeot18@verizon.invalid>
Wed, 24 Nov 2010 09:09:39 -0500
On 11/24/2010 04:29 AM, Sebastian wrote:

I am not sure how to understand that. Does it mean that removal from a
linked list, even through the remove method of an iterator over the
list, is implemented in terms of either the remove(int index)or
the remove(Object o) method? Which in a LinkedList would need to
traverse the list, making removal a linear time operation.

What it means is that it does the right thing when you try to use it
like this:

List<Integer> list =
   new LinkedList<Integer>(Arrays.asList(1, 2, 3, 4, 5));
Iterator<Integer> it = list.iterator();
while (it.hasNext()) {
   // Remove all even elements
   if ( % 2 == 0)

The iterator for LinkedLists happens to be implemented in terms of
having a pointer to a node in the linked list, specifically the last one
retrieved [1]. Removing that node is then an O(1) operation.

The wording may be a bit complicated, but it's basically describing the
sanest implementation: the iterator is a pointer to the element.
Operations like remove() removes the element being pointed to; next()
moves to the next element(); etc.

[1] I'm not sure about this part, but it's the easiest way to fulfill
the contracts of ListIterator.

Am I missing something? Is the documentation simply wrong?

-- Sebastian

Beware of bugs in the above code; I have only proved it correct, not
tried it. -- Donald E. Knuth

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