Any implementation of Ordered Set in Java?

Every Set has an iterator(). A normal HashSet's iterator is quite random, a TreeSet does it by sort order, a LinkedHashSet iterator iterates by insert order.

You can't replace an element in a LinkedHashSet, however. You can remove one and add another, but the new element will not be in the place of the original. In a LinkedHashMap, you can replace a value for an existing key, and then the values will still be in the original order.

Also, you can't insert at a certain position.

Maybe you'd better use an ArrayList with an explicit check to avoid inserting duplicates.


Take a look at the Java standard API doc. Right next to LinkedHashMap, there is a LinkedHashSet. But note that the order in those is the insertion order, not the natural order of the elements. And you can only iterate in that order, not do random access (except by counting iteration steps).

There is also an interface SortedSet implemented by TreeSet and ConcurrentSkipListSet. Both allow iteration in the natural order of their elements or a Comparator, but not random access or insertion order.

For a data structure that has both efficient access by index and can efficiently implement the set criterium, you'd need a skip list, but there is no implementation with that functionality in the Java Standard API, though I am certain it's easy to find one on the internet.


TreeSet is ordered.

http://docs.oracle.com/javase/6/docs/api/java/util/TreeSet.html


Take a look at LinkedHashSet class

From Java doc:

Hash table and linked list implementation of the Set interface, with predictable iteration order. This implementation differs from HashSet in that it maintains a doubly-linked list running through all of its entries. This linked list defines the iteration ordering, which is the order in which elements were inserted into the set (insertion-order). Note that insertion order is not affected if an element is re-inserted into the set. (An element e is reinserted into a set s if s.add(e) is invoked when s.contains(e) would return true immediately prior to the invocation.).