Leetcode: Zigzag Iterator

Given two 1d vectors, implement an iterator to return their elements alternately.

For example, given two 1d vectors:

v1 = [1, 2]
v2 = [3, 4, 5, 6]
By calling next repeatedly until hasNext returns false, the order of elements returned by next should be: [1, 3, 2, 4, 5, 6].

Follow up: What if you are given k 1d vectors? How well can your code be extended to such cases?

Clarification for the follow up question - Update (2015-09-18):
The "Zigzag" order is not clearly defined and is ambiguous for k > 2 cases. If "Zigzag" does not look right to you, replace "Zigzag" with "Cyclic". For example, given the following input:

[1,2,3]
[4,5,6,7]
[8,9]
It should return [1,4,8,2,5,9,3,6,7].

Best Solution: O(N) This solution also works for K vectors

 Use a queue to store the iterators in different vectors. Every time we call next(), we pop an element from the list, and re-add it to the end to cycle through the lists if it is not empty

public class ZigzagIterator {
    Queue<Iterator> list;
    public ZigzagIterator(List<Integer> v1, List<Integer> v2) {
        list = new LinkedList<Iterator>();
        if(!v1.isEmpty()) list.add(v1.iterator());
        if(!v2.isEmpty()) list.add(v2.iterator());
    }

    public int next() {
        Iterator poll = list.poll();
        int result = (Integer)poll.next();
        if(poll.hasNext()) list.add(poll);
        return result;
    }

    public boolean hasNext() {
        return !list.isEmpty();
    }
}

之前做法： 参考http://segmentfault.com/a/1190000003786218

Q：如果输入是k个列表呢？
A：使用一个迭代器的列表来管理这些迭代器。用turns变量和取模来判断我们该取列表中的第几个迭代器。不同点在于，一个迭代器用完后，我们要将其从列表中移出，这样我们下次就不用再找这个空的迭代器了。同样，由于每用完一个迭代器后都要移出一个，turns变量也要相应的更新为该迭代器下标的上一个下标。如果迭代器列表为空，说明没有下一个了。

public class ZigzagIterator implements Iterator<Integer> {
    
    List<Iterator<Integer>> itlist;
    int turns;

    public ZigzagIterator(List<Iterator<Integer>> list) {
        this.itlist = new LinkedList<Iterator<Integer>>();
        // 将非空迭代器加入列表
        for(Iterator<Integer> it : list){
            if(it.hasNext()){
                itlist.add(it);
            }
        }
        turns = 0;
    }

    public Integer next() {
        if(!hasNext()){
            return 0;
        }
        Integer res = 0;
        // 算出本次使用的迭代器的下标
        int pos = turns % itlist.size();
        Iterator<Integer> curr = itlist.get(pos);
        res = curr.next();
        // 如果这个迭代器用完，就将其从列表中移出
        if(!curr.hasNext()){
            itlist.remove(pos);
            // turns变量更新为上一个下标
            turns = pos - 1;
        }
        turns++;
        return res;
    }

    public boolean hasNext() {
        return itlist.size() > 0;
    }
}