Zigzag Iterator

题目地址：
https://leetcode.com/problems/zigzag-iterator/#/description

题目：

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].

解题思路：
这道题就是用一个iterator的list来存所有的iterator，然后next的时候每次remove第一个，然后放到list的最后。这道题用linkedlist比较好。

代码：

public class ZigzagIterator {

    List<Iterator> list;

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

    public int next() {
        Iterator iterator = list.remove(0);
        int rst = (int) iterator.next();
        if(iterator.hasNext()){
            list.add(iterator);
        }
        return rst;
    }

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

}