Question

我创建了一个方法，该方法应交织两个列表对象并返回新的交织后列表。

即。如果aList是[A，C，E，G]＆amp; bList是[B，D，F]，该方法将返回一个包含[A，B，C，D，E，F，G]的列表

没有编译器错误，当我运行程序时，返回的List为空。我已经发现该程序没有进入for循环，因为出于某种原因，当我创建 newList 时，它的大小为0.

以下是代码：

    public static <E> List<E> interweave(List<E> a, List<E> b){
    List<E> newList = new ArrayList<E>(a.size() + b.size());

    Iterator<E> itrA = a.iterator();
    Iterator<E> itrB = b.iterator();

    for(int i = 0; i < newList.size(); i++){
        if(i%2 == 0)
            newList.add(i, itrA.next());
        else
            newList.add(i, itrB.next());
    }

    return newList;
}

Answer 1

我相信这可以通过while(iterator.hasNext())惯用语：

以更简单的方式完成

itrA = a.iterator();
itrB = b.iterator(); 

while (itrA.hasNext() || itrB.hasNext()) {
  if (itrA.hasNext())
      newList.add(itrA.next());
  if (itrB.hasNext())
      newList.add(itrB.next());
}

Answer 2

可以动态组合列表：

static class InterWeave<T> implements Iterable<T> {

    // The lists to be interwoven.
    final Iterable<T>[] lists;

    public InterWeave(Iterable<T>... lists) {
        this.lists = lists;
    }

    @Override
    public Iterator<T> iterator() {
        return new WeaveIterator();

    }

    private class WeaveIterator implements Iterator<T> {

        // Which list to offer next.
        int whichList = 0;
        // The Iterators of those lists.
        final List<Iterator<T>> iterators = new ArrayList(lists.length);
        // The current iterator.
        Iterator<T> i;
        // The next to deliver - null menas not primed yet.
        T next = null;

        public WeaveIterator() {
            // Take some iterators.
            for (Iterable<T> l : lists) {
                if (l != null) {
                    iterators.add(l.iterator());
                }
            }
        }

        @Override
        public boolean hasNext() {
            // Have we already prepared one?
            if (next == null) {
                // Grab the next iterator and step on.
                i = iterators.get(whichList++);
                // detect that we're back at start.
                Iterator<T> start = i;
                // Cycle around the iterators.
                whichList %= iterators.size();
                while (!i.hasNext()) {
                    // Get next one.
                    i = iterators.get(whichList++);
                    whichList %= iterators.size();
                    if (i == start) {
                        // Stop here if we've gone all the way around.
                        break;
                    }
                }
                if (i.hasNext()) {
                    next = i.next();
                }
            }
            return next != null;
        }

        @Override
        public T next() {
            if (hasNext()) {
                T n = next;
                // Mark it used.
                next = null;
                return n;
            } else {
                throw new NoSuchElementException();
            }
        }

    }

}

public void test() {
    List<String> a = Arrays.asList(new String[]{"A", "C", "E", "G"});
    List<String> b = Arrays.asList(new String[]{"B", "D", "F"});
    for (String s : new InterWeave<String>(a, b)) {
        System.out.println(s);
    }
}

方法错误 - 使用Iterator交织列表

2 个答案: