1

我在使用 inner 时遇到了困难Iterator。

private List<List<? extends HasWord>> sentences = new ArrayList<List<? extends HasWord>>(); 
private Iterator<String> wordIterator = new Words();
private class Words implements Iterator<String> {

    int currSentence = 0;
    int currWord = 0;

    @Override
    public boolean hasNext() {
        return currSentence != sentences.size() - 1 && currWord != sentences.get(currSentence).size() - 1;
    }

    @Override
    public String next() {
        String nextWord = sentences.get(currSentence).get(currWord).word();
        currSentence++;
        currWord++;

        return nextWord;
    }

    @Override
    public void remove() {
        throw new UnsupportedOperationException();          
    }

}

然后，我尝试迭代它：

for (String s : wordIterator) { //Error: Can only iterate over an array or an instance of java.lang.Iterable
            words.add(s);

但它不起作用。（请参阅有问题的行上注释的编译器错误）。我在这里做错了什么？

在工程笔记上，解决我的问题的正确方法是什么？我有一堆这种形式的循环：

    for (List<? extends HasWord> sent : sentences) {
        for (HasWord token : sent) {
            //do stuff
        }
        }

所以我决定一个Iterator会更干净。这是矫枉过正，还是有其他方法可以做到？

4

2 回答 2

5

有两个嵌套for循环来执行此操作从根本上没有错，但我认为这会更干净：

public class Words implements Iterator<String> {
  private final Iterator<HasWord> sentences;
  private Iterator<String> currentSentence;

  public boolean hasNext() {
    return currentSentence.hasNext() || sentences.hasNext();
  }

  public String next() {
    if (currentSentence.hasNext()) {
      return currentSentence.next();
    }
    currentSentence = sentences.next();
    return next(); // will return the first word of the next sentence
  }
  //remove() omitted for brevity
}

每次你需要一个迭代多个句子时，返回这个类的一个新实例，并sentences使用初始化字段sentences.iterator();

（更仔细阅读您的问题后编辑）

于 2009-12-08T18:48:07.363 回答
3 
private class Words implements Iterator<String>, Iterable<String> {
  ...
  public Iterator<String> iterator() {
    return this;
  }
  ...
}
于 2009-12-08T20:51:18.377 回答