281. Zigzag Iterator

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].

思路: 用一个queue来实现, 把元素按照Cyclic的顺序放到queue里面.

public class ZigzagIterator {
    private ArrayDeque<Integer> queue;

    public ZigzagIterator(List<Integer> v1, List<Integer> v2) {
        queue = new ArrayDeque<>();
        this.addToQueue(v1, v2, queue);
    }

    public int next() {
        return queue.poll();
    }

    public boolean hasNext() {
        return !queue.isEmpty();
    }
    private void addToQueue(List<Integer> v1, List<Integer> v2, ArrayDeque<Integer> queue) {
        List<List<Integer>> list = new ArrayList<>();
        list.add(v1);
        list.add(v2);
        int size = Math.max(v1.size(), v2.size());
        for (int i = 0; i < size; i++) {
            for (int j = 0; j < list.size(); j++) { 
                if (list.get(j).size() <= i) continue; // 排除小于size的list
                Integer value = list.get(j).get(i);
                if (value != null) {
                    queue.offer(value);
                }
            }
        }

    }
}

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