LeetCode Zigzag Iterator

原题链接在这里：https://leetcode.com/problems/zigzag-iterator/

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].

题解：

用queue来保存每个list的iterator. next() 是 poll 出que的第一个iterator, return iterator.next(), 若是该iterator还有next, 就再放到queue尾部.

若queue空了，说明都遍历过了，此时hasNext()就返回false.

Time Complexity: next, O(1). hasNext, O(1). Space: O(l), l 是list的个数.

AC Java:

 public class ZigzagIterator {
     LinkedList<Iterator<Integer>> que;
     public ZigzagIterator(List<Integer> v1, List<Integer> v2) {
         que = new LinkedList<Iterator<Integer>>();
         if(v1.iterator().hasNext()){
             que.add(v1.iterator());
         }
         if(v2.iterator().hasNext()){
             que.add(v2.iterator());
         }
     }

     public int next() {
         Iterator<Integer> it = que.poll();
         int cur = it.next();
         if(it.hasNext()){
             que.add(it);
         }
         return cur;
     }

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

 /**
  * Your ZigzagIterator object will be instantiated and called as such:
  * ZigzagIterator i = new ZigzagIterator(v1, v2);
  * while (i.hasNext()) v[f()] = i.next();
  */

类似Flatten 2D Vector.