How about reversing the Collection backing the stream prior?

import java.util.Collections;
import java.util.List;

public void reverseTest(List<Integer> sampleCollection) {
    Collections.reverse(sampleCollection); // remember this reverses the elements in the list, so if you want the original input collection to remain untouched clone it first.

    sampleCollection.stream().forEach(item -> {
      // you op here
    });
}

List newStream = list.stream().sorted(Collections.reverseOrder()).collect(Collectors.toList());
        newStream.forEach(System.out::println);

一般问题:

流不存储任何元素。

因此，如果不将元素存储在某个中间集合中，就不可能以相反的顺序迭代元素。

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

更新:改变LinkedList为ArrayDeque(更好)，详情请看这里

打印:

3

20

2

1

顺便说一下，使用sort方法是不正确的，因为它排序，而不是反转(假设流可能有无序元素)

具体问题:

我发现这很简单，更容易和直观(复制@Holger评论)

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

你可以定义自己的收集器，按相反的顺序收集元素:

public static <T> Collector<T, List<T>, List<T>> inReverse() {
    return Collector.of(
        ArrayList::new,
        (l, t) -> l.add(t),
        (l, r) -> {l.addAll(r); return l;},
        Lists::<T>reverse);
}

像这样使用它:

stream.collect(inReverse()).forEach(t -> ...)

我使用ArrayList在前向顺序有效地插入收集项(在列表的末尾)，和番石榴列表。Reverse用于有效地提供列表的反向视图，而无需复制另一个列表。

下面是自定义收集器的一些测试用例:

import static org.hamcrest.MatcherAssert.assertThat;
import static org.hamcrest.Matchers.*;

import java.util.ArrayList;
import java.util.List;
import java.util.function.BiConsumer;
import java.util.function.BinaryOperator;
import java.util.function.Function;
import java.util.function.Supplier;
import java.util.stream.Collector;

import org.hamcrest.Matchers;
import org.junit.Test;

import com.google.common.collect.Lists;

public class TestReverseCollector {
    private final Object t1 = new Object();
    private final Object t2 = new Object();
    private final Object t3 = new Object();
    private final Object t4 = new Object();

    private final Collector<Object, List<Object>, List<Object>> inReverse = inReverse();
    private final Supplier<List<Object>> supplier = inReverse.supplier();
    private final BiConsumer<List<Object>, Object> accumulator = inReverse.accumulator();
    private final Function<List<Object>, List<Object>> finisher = inReverse.finisher();
    private final BinaryOperator<List<Object>> combiner = inReverse.combiner();

    @Test public void associative() {
        final List<Object> a1 = supplier.get();
        accumulator.accept(a1, t1);
        accumulator.accept(a1, t2);
        final List<Object> r1 = finisher.apply(a1);

        final List<Object> a2 = supplier.get();
        accumulator.accept(a2, t1);
        final List<Object> a3 = supplier.get();
        accumulator.accept(a3, t2);
        final List<Object> r2 = finisher.apply(combiner.apply(a2, a3));

        assertThat(r1, Matchers.equalTo(r2));
    }

    @Test public void identity() {
        final List<Object> a1 = supplier.get();
        accumulator.accept(a1, t1);
        accumulator.accept(a1, t2);
        final List<Object> r1 = finisher.apply(a1);

        final List<Object> a2 = supplier.get();
        accumulator.accept(a2, t1);
        accumulator.accept(a2, t2);
        final List<Object> r2 = finisher.apply(combiner.apply(a2, supplier.get()));

        assertThat(r1, equalTo(r2));
    }

    @Test public void reversing() throws Exception {
        final List<Object> a2 = supplier.get();
        accumulator.accept(a2, t1);
        accumulator.accept(a2, t2);

        final List<Object> a3 = supplier.get();
        accumulator.accept(a3, t3);
        accumulator.accept(a3, t4);

        final List<Object> r2 = finisher.apply(combiner.apply(a2, a3));

        assertThat(r2, contains(t4, t3, t2, t1));
    }

    public static <T> Collector<T, List<T>, List<T>> inReverse() {
        return Collector.of(
            ArrayList::new,
            (l, t) -> l.add(t),
            (l, r) -> {l.addAll(r); return l;},
            Lists::<T>reverse);
    }
}

反转流的正确通用方法是什么?

如果流没有指定遇到顺序，就不要这样做。 (! s.spliterator () .hasCharacteristics (java.util.Spliterator.ORDERED))

最简单的解决方案是使用List::listIterator和Stream::generate

List<Integer> list = Arrays.asList(1, 2, 3, 4, 5);
ListIterator<Integer> listIterator = list.listIterator(list.size());

Stream.generate(listIterator::previous)
      .limit(list.size())
      .forEach(System.out::println);