如何为2D数组/列表创建Java迭代器

Question

最近有人问我一个问题，即如何为2D数组创建Java迭代器，特别是如何实现：

public class PersonIterator implements Iterator<Person>{
    private List<List<Person>> list;

    public PersonIterator(List<List<Person>> list){
        this.list = list;
    }

    @Override
    public boolean hasNext() {
    }

    @Override
    public Person next() {

    }
}

一维数组通过使用索引来跟踪位置非常简单明了，有关如何对2D列表执行此操作的任何想法。

Answer 1

在1D情况下，您需要保留一个索引来知道您离开的位置，对吗？ 好吧，在2D情况下，您需要两个索引：一个索引要知道您在哪个子列表中工作，另一个索引要知道您在该子列表中的哪个元素上留下了。

Answer 2

像这样吗 （注：未经测试）

public class PersonIterator implements Iterator<Person>{
    // This keeps track of the outer set of lists, the lists of lists
    private Iterator<List<Person>> iterator;
    // This tracks the inner set of lists, the lists of persons we're going through
    private Iterator<Person> curIterator;

    public PersonIterator(List<List<Person>> list){
        // Set the outer one
        this.iterator = list.iterator();
        // And set the inner one based on whether or not we can
        if (this.iterator.hasNext()) {
            this.curIterator = iterator.next();
        } else {
            this.curIterator = null;
        }
    }

    @Override
    public boolean hasNext() {
         // If the current iterator is valid then we obviously have another one
         if (curIterator != null && curIterator.hasNext()) {
             return true;
         // Otherwise we need to safely get the iterator for the next list to iterate.
         } else if (iterator.hasNext()) {
             // We load a new iterator here
             curIterator = iterator.next();
             // and retry peeking to see if the new curIterator has any elements to iterate.
             return hasNext();
         // Otherwise we're out of lists.
         } else {
             return false;
         }
    }

    @Override
    public Person next() {
         // Return the current value off the inner iterator if we can
         if (curIterator != null && curIterator.hasNext()) {
             return curIterator.next();
         // Otherwise try to iterate along the next list and retry getting the next one.
         // This won't infinitely loop at the end since next() at the end of the outer
         // iterator should result in an NoSuchElementException.
         } else {
             curIterator = iterator.next();
             return next();
         }
    }
}