General question: What's the proper way to reverse a stream? Assuming that we don't know what type of elements that stream consists of, what's the generic way to reverse any stream?

Specific question:

IntStream provides range method to generate Integers in specific range IntStream.range(-range, 0), now that I want to reverse it switching range from 0 to negative won't work, also I can't use Integer::compare

List<Integer> list = Arrays.asList(1,2,3,4);
list.stream().sorted(Integer::compare).forEach(System.out::println);

with IntStream I'll get this compiler error

Error:(191, 0) ajc: The method sorted() in the type IntStream is not applicable for the arguments (Integer::compare)

what am I missing here?


For the specific question of generating a reverse IntStream, try something like this:

static IntStream revRange(int from, int to) {
    return IntStream.range(from, to)
                    .map(i -> to - i + from - 1);
}

This avoids boxing and sorting.

For the general question of how to reverse a stream of any type, I don't know of there's a "proper" way. There are a couple ways I can think of. Both end up storing the stream elements. I don't know of a way to reverse a stream without storing the elements.

This first way stores the elements into an array and reads them out to a stream in reverse order. Note that since we don't know the runtime type of the stream elements, we can't type the array properly, requiring an unchecked cast.

@SuppressWarnings("unchecked")
static <T> Stream<T> reverse(Stream<T> input) {
    Object[] temp = input.toArray();
    return (Stream<T>) IntStream.range(0, temp.length)
                                .mapToObj(i -> temp[temp.length - i - 1]);
}

Another technique uses collectors to accumulate the items into a reversed list. This does lots of insertions at the front of ArrayList objects, so there's lots of copying going on.

Stream<T> input = ... ;
List<T> output =
    input.collect(ArrayList::new,
                  (list, e) -> list.add(0, e),
                  (list1, list2) -> list1.addAll(0, list2));

It's probably possible to write a much more efficient reversing collector using some kind of customized data structure.

UPDATE 2016-01-29

Since this question has gotten a bit of attention recently, I figure I should update my answer to solve the problem with inserting at the front of ArrayList. This will be horribly inefficient with a large number of elements, requiring O(N^2) copying.

It's preferable to use an ArrayDeque instead, which efficiently supports insertion at the front. A small wrinkle is that we can't use the three-arg form of Stream.collect(); it requires the contents of the second arg be merged into the first arg, and there's no "add-all-at-front" bulk operation on Deque. Instead, we use addAll() to append the contents of the first arg to the end of the second, and then we return the second. This requires using the Collector.of() factory method.

The complete code is this:

Deque<String> output =
    input.collect(Collector.of(
        ArrayDeque::new,
        (deq, t) -> deq.addFirst(t),
        (d1, d2) -> { d2.addAll(d1); return d2; }));

The result is a Deque instead of a List, but that shouldn't be much of an issue, as it can easily be iterated or streamed in the now-reversed order.


Elegant solution

List<Integer> list = Arrays.asList(1,2,3,4);
list.stream()
    .sorted(Collections.reverseOrder()) // Method on Stream<Integer>
    .forEach(System.out::println);

General Question:

Stream does not store any elements.

So iterating elements in the reverse order is not possible without storing the elements in some intermediate collection.

Stream.of("1", "2", "20", "3")
      .collect(Collectors.toCollection(ArrayDeque::new)) // or LinkedList
      .descendingIterator()
      .forEachRemaining(System.out::println);

Update: Changed LinkedList to ArrayDeque (better) see here for details

Prints:

3

20

2

1

By the way, using sort method is not correct as it sorts, NOT reverses (assuming stream may have unordered elements)

Specific Question:

I found this simple, easier and intuitive(Copied @Holger comment)

IntStream.iterate(to - 1, i -> i - 1).limit(to - from)

Many of the solutions here sort or reverse the IntStream, but that unnecessarily requires intermediate storage. Stuart Marks's solution is the way to go:

static IntStream revRange(int from, int to) {
    return IntStream.range(from, to).map(i -> to - i + from - 1);
}

It correctly handles overflow as well, passing this test:

@Test
public void testRevRange() {
    assertArrayEquals(revRange(0, 5).toArray(), new int[]{4, 3, 2, 1, 0});
    assertArrayEquals(revRange(-5, 0).toArray(), new int[]{-1, -2, -3, -4, -5});
    assertArrayEquals(revRange(1, 4).toArray(), new int[]{3, 2, 1});
    assertArrayEquals(revRange(0, 0).toArray(), new int[0]);
    assertArrayEquals(revRange(0, -1).toArray(), new int[0]);
    assertArrayEquals(revRange(MIN_VALUE, MIN_VALUE).toArray(), new int[0]);
    assertArrayEquals(revRange(MAX_VALUE, MAX_VALUE).toArray(), new int[0]);
    assertArrayEquals(revRange(MIN_VALUE, MIN_VALUE + 1).toArray(), new int[]{MIN_VALUE});
    assertArrayEquals(revRange(MAX_VALUE - 1, MAX_VALUE).toArray(), new int[]{MAX_VALUE - 1});
}

without external lib...

import java.util.List;
import java.util.Collections;
import java.util.stream.Collector;

public class MyCollectors {

    public static <T> Collector<T, ?, List<T>> toListReversed() {
        return Collectors.collectingAndThen(Collectors.toList(), l -> {
            Collections.reverse(l);
            return l;
        });
    }

}