281. Zigzag Iterator | Keep Walking

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

For example, given two 1d vectors:

1 2	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

[1,2,3]
[4,5,6,7]
[8,9]

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

直接做Follow Up吧。也比较简单，但是直接在Queue里面放Iterator的做法还是没有想到。

下面是我找到的比较好的代码：

public class ZigzagIterator {
    
    Queue<Iterator> q;
    public ZigzagIterator(List<Integer> v1, List<Integer> v2) {
        q = new LinkedList();
        if (!v1.isEmpty()) q.offer(v1.iterator());
        if (!v2.isEmpty()) q.offer(v2.iterator());
    }
    public int next() {
        Iterator cur = q.poll();
        int res = (int) cur.next();
        if (cur.hasNext()) q.offer(cur);
        return res;
    }
    public boolean hasNext() {
        return !q.isEmpty(); 
    }
}

这是我自己的做法，比较慢：

public class ZigzagIterator {
    Queue<Integer> q = new LinkedList<>();
    public ZigzagIterator(List<Integer> v1, List<Integer> v2) {
        List<List<Integer>> list = new ArrayList<>();
        list.add(v1);
        list.add(v2);
        for (int i = 0; list.size() != 0; i++) {
            Iterator<List<Integer>> iter = list.iterator();
            while (iter.hasNext()) {
                List<Integer> v = iter.next();
                if (v.size() <= i)
                    iter.remove();
                else
                    q.add(v.get(i));
            }
        }
    }
    public int next() {
        return q.poll();
    }
    public boolean hasNext() {
        return !q.isEmpty();
    }
}

Ref: