For a finite range, you can iterate over the range while
maintaining a collection of the top n elements and discarding
duplicates as you find them. For example, using a
RedBlackTree as the collection, you will get O(log(n) *
range.length) total time and O(n) total memory used. If n is
small (on the order of 10s), an array will suffice, resulting
in O(n * range.length) total time but with lower constant
factor.
You can also use an array if you use a BinaryHeap. If you use a
heap then it should have a much lower constant factor than a
red-black tree so, even though they would both have O(m*log(n))
(where n is the number of elements you want the top of and m is
the size of the collection) running time, a heap would be
faster.
My first thought was about a heap, too, but it does not allow to
search for a duplicate in O(log(n)) as it guarantees only partial
ordering.
