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

Example:

Input:v1 = [1,2]v2 = [3,4,5,6] Output: [1,3,2,4,5,6]Explanation: 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:

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:

Input:[1,2,3][4,5,6,7][8,9]Output: [1,4,8,2,5,9,3,6,7].

 

用一个queue来存每个list的iterator。call next()时，从q中poll出先进queue的iterator，返回其next值，如果它没到末尾，还要把它存回queue以备下次使用。call hasNext()就是判断queue是否为空。

时间：O(N)，空间：O(1)

public class ZigzagIterator {    Queue
     
       q;    public ZigzagIterator(List
      
        v1, List
       
         v2) {        q = new LinkedList<>();        if(!v1.isEmpty())            q.offer(v1.iterator());        if(!v2.isEmpty())            q.offer(v2.iterator());    }    public int next() {        Iterator iter = q.poll();        int val = (int)iter.next();        if(iter.hasNext())            q.offer(iter);        return val;    }    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(); */