Java：SortedSet“光标”样式的迭代器

Question

I need to iterate both forwards and backwards in a sorted set. If I use NavigableSet , I get a strictly-forward iterator and a strictly-backward iterator ( iterator() and descendingIterator() ) but none that can move forward and backward.

What's the time complexity of NavigableSet.lower() and higher() ? I can use those instead, but am reluctant to do so if they are inefficient.

Answer 1

Depending on your exact needs you could convert the sorted set to a list, say an array list, and use a list iterator for traversal. It can be used in both directions via the next() and previous() methods, which may be mixed freely.

Answer 2

There are only two implementations of the NavigableSet. Saying you opted for the TreeSet, while I don't have the source handy, the Javadoc says that it is based on a TreeMap providing O(log(n)) for get/put/containsKey/remove . At worst this would perform one get to find the value of we're finding the lower/higher for, plus an additional search to get the next/previous value, providing O(2log(n)) = O(log(n)).

Trees are worst case O(n) for search in the event it is actually a list, but in general, O(height).