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281. Zigzag Iterator



Given two 1d vectors, implement an iterator to return their elements alternately.
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:
Output: [1,4,8,2,5,9,3,6,7].


Approach #1

public class ZigzagIterator {
LinkedList<Iterator> list;
public ZigzagIterator(List<Integer> v1, List<Integer> v2) {
list = new LinkedList<Iterator>();
if (!v1.isEmpty()) list.add(v1.iterator());
if (!v2.isEmpty()) list.add(v2.iterator());
public int next() {
Iterator poll = list.remove(); // 头
int result = (Integer);
if (poll.hasNext) list.add(poll);
return result;
public boolean hasNext() {
return !list.isEmpty();
* Your ZigzagIterator object will be instantiated and called as such:
* ZigzagIterator i = new ZigzagIterator(v1, v2);
* while (i.hasNext()) v[f()] =;