如何为2D数组/列表创建Java迭代器

Question

I was recently asked about the question that how to create a Java Iterator for 2D Array, specifically how to implement:

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() {

    }
}

1D array is pretty straightforward by using a index to track the position, any idea about how to do it for 2D lists.

Answer 1

In the 1D case you need to keep one index to know where you left, right? Well, in the 2D case you need two indices: one to know in which sub-list you were working, and other one to know at what element inside that sub-list you left.

Answer 2

Something like this? (Note: untested)

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();
         }
    }
}