> For the complete documentation index, see [llms.txt](https://hao-fu-1.gitbook.io/oj/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://hao-fu-1.gitbook.io/oj/multiple-pointers/zigzag-iterator.md).

# Zigzag Iterator

<https://leetcode.com/problems/zigzag-iterator/description/>

> 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].

## Thoughts

循环的依次遍历, 也就是每次遍历过的排到队尾.

## Code

```
public class ZigzagIterator {
    Queue<Iterator<Integer>> iters;

    public ZigzagIterator(List<Integer> v1, List<Integer> v2) {
        iters = new LinkedList<>();
        if (!v1.isEmpty()) {
            iters.offer(v1.iterator());
        }
        if (!v2.isEmpty()) {
            iters.offer(v2.iterator());
        }
    }

    public int next() {
        Iterator<Integer> iter = iters.poll();
        int res = iter.next();
        if (iter.hasNext()) {
            iters.offer(iter);
        }
        return res;
    }

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

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

## Analysis

时间复杂度O(N), N为所有节点数目.