281. Zigzag Iterator - Medium

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<Iterator> q;

    public ZigzagIterator(List<Integer> v1, List<Integer> 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();
 */

原文地址：https://www.cnblogs.com/fatttcat/p/10037506.html