你可以使用IntStream.iterate()来获取索引:

String[] names = {"Sam","Pamela", "Dave", "Pascal", "Erik"};
List<String> nameList = IntStream.iterate(0, i -> i < names.length, i -> i + 1)
        .filter(i -> names[i].length() <= i)
        .mapToObj(i -> names[i])
        .collect(Collectors.toList());

这只适用于Java 9以上的Java 8，你可以使用这个:

String[] names = {"Sam","Pamela", "Dave", "Pascal", "Erik"};
List<String> nameList = IntStream.iterate(0, i -> i + 1)
        .limit(names.length)
        .filter(i -> names[i].length() <= i)
        .mapToObj(i -> names[i])
        .collect(Collectors.toList());

我在我的项目中使用了以下解决方案。我认为这比使用可变对象或整数范围要好。

import java.util.*;
import java.util.function.*;
import java.util.stream.Collector;
import java.util.stream.Collector.Characteristics;
import java.util.stream.Stream;
import java.util.stream.StreamSupport;
import static java.util.Objects.requireNonNull;


public class CollectionUtils {
    private CollectionUtils() { }

    /**
     * Converts an {@link java.util.Iterator} to {@link java.util.stream.Stream}.
     */
    public static <T> Stream<T> iterate(Iterator<? extends T> iterator) {
        int characteristics = Spliterator.ORDERED | Spliterator.IMMUTABLE;
        return StreamSupport.stream(Spliterators.spliteratorUnknownSize(iterator, characteristics), false);
    }

    /**
     * Zips the specified stream with its indices.
     */
    public static <T> Stream<Map.Entry<Integer, T>> zipWithIndex(Stream<? extends T> stream) {
        return iterate(new Iterator<Map.Entry<Integer, T>>() {
            private final Iterator<? extends T> streamIterator = stream.iterator();
            private int index = 0;

            @Override
            public boolean hasNext() {
                return streamIterator.hasNext();
            }

            @Override
            public Map.Entry<Integer, T> next() {
                return new AbstractMap.SimpleImmutableEntry<>(index++, streamIterator.next());
            }
        });
    }

    /**
     * Returns a stream consisting of the results of applying the given two-arguments function to the elements of this stream.
     * The first argument of the function is the element index and the second one - the element value. 
     */
    public static <T, R> Stream<R> mapWithIndex(Stream<? extends T> stream, BiFunction<Integer, ? super T, ? extends R> mapper) {
        return zipWithIndex(stream).map(entry -> mapper.apply(entry.getKey(), entry.getValue()));
    }

    public static void main(String[] args) {
        String[] names = {"Sam", "Pamela", "Dave", "Pascal", "Erik"};

        System.out.println("Test zipWithIndex");
        zipWithIndex(Arrays.stream(names)).forEach(entry -> System.out.println(entry));

        System.out.println();
        System.out.println("Test mapWithIndex");
        mapWithIndex(Arrays.stream(names), (Integer index, String name) -> index+"="+name).forEach((String s) -> System.out.println(s));
    }
}

你可以创建一个静态的内部类来封装索引器，就像我在下面的例子中需要做的那样:

static class Indexer {
    int i = 0;
}

public static String getRegex() {
    EnumSet<MeasureUnit> range = EnumSet.allOf(MeasureUnit.class);
    StringBuilder sb = new StringBuilder();
    Indexer indexer = new Indexer();
    range.stream().forEach(
            measureUnit -> {
                sb.append(measureUnit.acronym);
                if (indexer.i < range.size() - 1)
                    sb.append("|");

                indexer.i++;
            }
    );
    return sb.toString();
}

使用列表，你可以尝试一下

List<String> strings = new ArrayList<>(Arrays.asList("First", "Second", "Third", "Fourth", "Fifth")); // An example list of Strings
strings.stream() // Turn the list into a Stream
    .collect(HashMap::new, (h, o) -> h.put(h.size(), o), (h, o) -> {}) // Create a map of the index to the object
        .forEach((i, o) -> { // Now we can use a BiConsumer forEach!
            System.out.println(String.format("%d => %s", i, o));
        });

输出:

0 => First
1 => Second
2 => Third
3 => Fourth
4 => Fifth

你可以使用IntStream.iterate()来获取索引:

String[] names = {"Sam","Pamela", "Dave", "Pascal", "Erik"};
List<String> nameList = IntStream.iterate(0, i -> i < names.length, i -> i + 1)
        .filter(i -> names[i].length() <= i)
        .mapToObj(i -> names[i])
        .collect(Collectors.toList());

这只适用于Java 9以上的Java 8，你可以使用这个:

String[] names = {"Sam","Pamela", "Dave", "Pascal", "Erik"};
List<String> nameList = IntStream.iterate(0, i -> i + 1)
        .limit(names.length)
        .filter(i -> names[i].length() <= i)
        .mapToObj(i -> names[i])
        .collect(Collectors.toList());

我在这里找到了解决方案，当流创建的列表或数组(你知道的大小)。但是如果Stream的大小未知呢?在这种情况下，试试这个变体:

public class WithIndex<T> {
    private int index;
    private T value;

    WithIndex(int index, T value) {
        this.index = index;
        this.value = value;
    }

    public int index() {
        return index;
    }

    public T value() {
        return value;
    }

    @Override
    public String toString() {
        return value + "(" + index + ")";
    }

    public static <T> Function<T, WithIndex<T>> indexed() {
        return new Function<T, WithIndex<T>>() {
            int index = 0;
            @Override
            public WithIndex<T> apply(T t) {
                return new WithIndex<>(index++, t);
            }
        };
    }
}

用法:

public static void main(String[] args) {
    Stream<String> stream = Stream.of("a", "b", "c", "d", "e");
    stream.map(WithIndex.indexed()).forEachOrdered(e -> {
        System.out.println(e.index() + " -> " + e.value());
    });
}