将多个集合合并为一个逻辑集合？

Question

假设,我有一个恒定数量的集合(例如3个ArrayLists)作为类的成员.现在,我想将所有元素公开给其他类,以便它们可以简单地遍历所有元素(理想情况下,只读).我正在使用guava集合,我想知道如何使用guava iterables/iterators来生成内部集合的逻辑视图,而无需制作临时副本.

Answer 1

使用Guava,您可以使用Iterables.concat(Iterable<T> ...)它创建所有迭代的实时视图,连接成一个(如果您更改迭代,连接版本也会更改).然后将连接的iterable包装起来Iterables.unmodifiableIterable(Iterable<T>)(我之前没有看过只读的要求).

来自Iterables.concat( .. )JavaDocs:

将多个迭代组合成一个可迭代的.返回的iterable有一个迭代器,遍历输入中每个iterable的元素.必要时不会轮询输入迭代器.返回的iterable的迭代器支持remove() 相应的输入迭代器支持它.

虽然这没有明确说明这是一个实时视图,但最后一句暗示它是(Iterator.remove()仅当支持迭代器支持它时才支持该方法,除非使用实时视图)

示例代码:

final List<Integer> first  = Lists.newArrayList(1, 2, 3);
final List<Integer> second = Lists.newArrayList(4, 5, 6);
final List<Integer> third  = Lists.newArrayList(7, 8, 9);
final Iterable<Integer> all =
    Iterables.unmodifiableIterable(
        Iterables.concat(first, second, third));
System.out.println(all);
third.add(9999999);
System.out.println(all);

输出:

[1,2,3,4,5,6,7,8,9]
[1,2,3,4,5,6,7,8,9,9999999]

编辑:

根据Damian的请求,这是一个返回实时Collection View的类似方法

public final class CollectionsX {

    static class JoinedCollectionView<E> implements Collection<E> {

        private final Collection<? extends E>[] items;

        public JoinedCollectionView(final Collection<? extends E>[] items) {
            this.items = items;
        }

        @Override
        public boolean addAll(final Collection<? extends E> c) {
            throw new UnsupportedOperationException();
        }

        @Override
        public void clear() {
            for (final Collection<? extends E> coll : items) {
                coll.clear();
            }
        }

        @Override
        public boolean contains(final Object o) {
            throw new UnsupportedOperationException();
        }

        @Override
        public boolean containsAll(final Collection<?> c) {
            throw new UnsupportedOperationException();
        }

        @Override
        public boolean isEmpty() {
            return !iterator().hasNext();
        }

        @Override
        public Iterator<E> iterator() {
            return Iterables.concat(items).iterator();
        }

        @Override
        public boolean remove(final Object o) {
            throw new UnsupportedOperationException();
        }

        @Override
        public boolean removeAll(final Collection<?> c) {
            throw new UnsupportedOperationException();
        }

        @Override
        public boolean retainAll(final Collection<?> c) {
            throw new UnsupportedOperationException();
        }

        @Override
        public int size() {
            int ct = 0;
            for (final Collection<? extends E> coll : items) {
                ct += coll.size();
            }
            return ct;
        }

        @Override
        public Object[] toArray() {
            throw new UnsupportedOperationException();
        }

        @Override
        public <T> T[] toArray(T[] a) {
            throw new UnsupportedOperationException();
        }

        @Override
        public boolean add(E e) {
            throw new UnsupportedOperationException();
        }

    }

    /**
     * Returns a live aggregated collection view of the collections passed in.
     * <p>
     * All methods except {@link Collection#size()}, {@link Collection#clear()},
     * {@link Collection#isEmpty()} and {@link Iterable#iterator()}
     *  throw {@link UnsupportedOperationException} in the returned Collection.
     * <p>
     * None of the above methods is thread safe (nor would there be an easy way
     * of making them).
     */
    public static <T> Collection<T> combine(
        final Collection<? extends T>... items) {
        return new JoinedCollectionView<T>(items);
    }

    private CollectionsX() {
    }

}

Answer 2

简单的Java 8解决方案使用Stream.

常数

假设private Collection<T> c, c2, c3.

一个解决方案

public Stream<T> stream() {
    return Stream.concat(Stream.concat(c.stream(), c2.stream()), c3.stream());
}

另一种方案:

public Stream<T> stream() {
    return Stream.of(c, c2, c3).flatMap(Collection::stream);
}

可变数量

假设private Collection<Collection<T>> cs:

public Stream<T> stream() {
    return cs.stream().flatMap(Collection::stream);
}

Answer 3

如果您至少使用Java 8,请参阅我的其他答案.

如果您已经在使用Google Guava,请参阅Sean Patrick Floyd的回答.

如果你坚持在Java 7的和不希望包括谷歌番石榴,您可以在自己(只读)写Iterables.concat()不使用超过Iterable和Iterator:

常数

public static <E> Iterable<E> concat(final Iterable<? extends E> iterable1,
                                     final Iterable<? extends E> iterable2) {
    return new Iterable<E>() {
        @Override
        public Iterator<E> iterator() {
            return new Iterator<E>() {
                final Iterator<? extends E> iterator1 = iterable1.iterator();
                final Iterator<? extends E> iterator2 = iterable2.iterator();

                @Override
                public boolean hasNext() {
                    return iterator1.hasNext() || iterator2.hasNext();
                }

                @Override
                public E next() {
                    return iterator1.hasNext() ? iterator1.next() : iterator2.next();
                }
            };
        }
    };
}

可变数量

@SafeVarargs
public static <E> Iterable<E> concat(final Iterable<? extends E>... iterables) {
    return concat(Arrays.asList(iterables));
}

public static <E> Iterable<E> concat(final Iterable<Iterable<? extends E>> iterables) {
    return new Iterable<E>() {
        final Iterator<Iterable<? extends E>> iterablesIterator = iterables.iterator();

        @Override
        public Iterator<E> iterator() {
            return !iterablesIterator.hasNext() ? Collections.emptyIterator()
                                                : new Iterator<E>() {
                Iterator<? extends E> iterableIterator = nextIterator();

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

                @Override
                public E next() {
                    final E next = iterableIterator.next();
                    findNext();
                    return next;
                }

                Iterator<? extends E> nextIterator() {
                    return iterablesIterator.next().iterator();
                }

                Iterator<E> findNext() {
                    while (!iterableIterator.hasNext()) {
                        if (!iterablesIterator.hasNext()) {
                            break;
                        }
                        iterableIterator = nextIterator();
                    }
                    return this;
                }
            }.findNext();
        }
    };
}