Get last n elements from stream

A custom collector can be written like this:

public static <T> Collector<T, ?, List<T>> lastN(int n) {
    return Collector.<T, Deque<T>, List<T>>of(ArrayDeque::new, (acc, t) -> {
        if(acc.size() == n)
            acc.pollFirst();
        acc.add(t);
    }, (acc1, acc2) -> {
        while(acc2.size() < n && !acc1.isEmpty()) {
            acc2.addFirst(acc1.pollLast());
        }
        return acc2;
    }, ArrayList::new);
}

And use it like this:

List<String> lastTen = input.stream().collect(lastN(10));

Use Stream.skip()

Returns a stream consisting of the remaining elements of this stream after discarding the first n elements of the stream. If this stream contains fewer than n elements then an empty stream will be returned.

all.stream().skip(Math.max(0, all.size() - n)).forEach(doSomething);

In case the stream has unknown size, there's probably no way around consuming the entire stream and buffering the last n elements encountered so far. You can do this using some kind of deque, or a specialized ring-buffer automatically maintaining its maximum size (see this related question for some implementations).

public static <T> List<T> lastN(Stream<T> stream, int n) {
    Deque<T> result = new ArrayDeque<>(n);
    stream.forEachOrdered(x -> {
        if (result.size() == n) {
            result.pop();
        }
        result.add(x);
    });
    return new ArrayList<>(result);
}

All of those operations (size, pop, add) should have complexity of O(1), so the overall complexity for a stream with (unknown) length n would be O(n).

Sometimes I need a "oneliner" (in this case a three liner) as creating a collector is just too much fuss.

If the stream is small then it is possible to reverse, limit and reverse again without much sacrificing performance. This will result the last n elements.

It is useful if filtering is required as in that case it is not possible to specify the size.

Stream.of(1, 2, 3, 4, 5, 6, 7, 8, 9)
  .filter(i -> i % 2 == 0)
  .sorted(Comparator.reverseOrder())
  .limit(2)
  .sorted(Comparator.naturalOrder())
  .forEach(System.out::println); // prints 6 8