Zigzag Iterator
題意：
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].
解題思路：
使用一個q來維護所有的iterator，一但一個iterator被存取過了，他會被poll出來並檢查是否還有下一個元素，若有的話，則往q的後面塞回去。
public class ZigzagIterator {
private Queue<Iterator<Integer>> q;
public ZigzagIterator(List<Integer> v1, List<Integer> v2) {
q = new LinkedList<Iterator<Integer>>();
if (!v1.isEmpty()) {
q.offer(v1.iterator());
}
if (!v2.isEmpty()) {
q.offer(v2.iterator());
}
}
public int next() {
Iterator cur = q.poll();
int res = (Integer) cur.next();
if (cur.hasNext()) {
q.offer(cur);
}
return res;
}
public boolean hasNext() {
return !q.isEmpty();
}
}
/**
* Your ZigzagIterator object will be instantiated and called as such:
* ZigzagIterator i = new ZigzagIterator(v1, v2);
* while (i.hasNext()) v[f()] = i.next();
*/