java – 将多个Iterables包装到一个Interable中

发布时间：2020-12-15 08:39:10 所属栏目：Java 来源：网络整理

导读：说我有两个收藏： Collection Integer foo = new ArrayList Integer ();Collection Integer bar = new ArrayList Integer (); 并说有时我想单独迭代它们,但有时一起.有没有办法在foo和bar周围创建一个包装器,这样我就可以迭代组合对,但每当foo和bar改变时它

说我有两个收藏：

Collection< Integer > foo = new ArrayList< Integer >();
Collection< Integer > bar = new ArrayList< Integer >();

并说有时我想单独迭代它们,但有时一起.有没有办法在foo和bar周围创建一个包装器,这样我就可以迭代组合对,但每当foo和bar改变时它也会更新？ (即Collection.addAll()不适合).

例如：

Collection< Integer > wrapper = ... // holds references to both bar and foo

foo.add( 1 );
bar.add( 99 );

for( Integer fooInt : foo ) {
    System.out.println( fooInt );
} // output: 1

for( Integer barInt : bar ) {
    System.out.println( barInt );
} // output: 99

for( Integer wrapInt : wrapper ) {
    System.out.println( wrapInt );
} // output: 1,99

foo.add( 543 );

for( Integer wrapInt : wrapper ) {
    System.out.println( wrapInt );
} // output: 1,99,543

谢谢！

解决方法

我写了两个函数：

/**
 * Create an Iterator from multiple Iterators. The returned Iterator
 * traverses all elements from all sources,in the order,as if they belong
 * to the same source.
 * 
 * @param <T> type of elements
 * @param sources sources of the elements,in order of traversal
 * @return an iterator over multiple iterators in sequence
 */
public static <T> Iterator<T> concatenate(final Iterator<T> ... sources) {
    if (sources.length == 0) {
        return new Iterator<T>() {
            @Override public boolean hasNext() { return false; }
            @Override public T next() { throw new NoSuchElementException("end of iteration"); }
            @Override public void remove() { throw new IllegalStateException("no previous element"); }
        };
    }
    return new Iterator<T>() {

        Iterator<Iterator<T>> sourcesIterator = Arrays.asList(sources).iterator();
        Iterator<T> currentIterator = sourcesIterator.next();

        @Override
        public boolean hasNext() {
            if (currentIterator.hasNext()) {
                return true;
            } else {
                if (sourcesIterator.hasNext()) {
                    currentIterator = sourcesIterator.next();
                    return hasNext();
                } else {
                    return false;
                }
            }
        }

        @Override
        public T next() {
            if (hasNext()) {
                return currentIterator.next();
            } else {
                throw new NoSuchElementException("end of iteration");
            }
        }

        @Override
        public void remove() {
            currentIterator.remove();
        }
    };
}

/**
 * Create an Iterable from multiple Iterables. The returned Iterable
 * traverses all elements from all sources,in order of traversal
 * @return an iterable over multiple iterators in sequence
 */
@SuppressWarnings("unchecked") // impossible to create a generic array
public static <T> Iterable<T> concatenate(final Iterable<T> ... sources) {
    return new Iterable<T>() {
        @Override
        public Iterator<T> iterator() {
            final Iterator[] iteratorsArrays = new Iterator[sources.length];
            for (int i = 0; i < sources.length; i++) {
                iteratorsArrays[i] = sources[i].iterator();
            }
            return concatenate(iteratorsArrays);
        }
    };
}

（编辑：李大同）

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