Zigzag Iterator

Total Accepted: 7833 Total Submissions: 18659 Difficulty: Medium
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

• 考察Iterator
• 利用其本身的iterator

public class ZigzagIterator {
    Iterator<Integer>[] its;
    int pos;

    public ZigzagIterator(List<Integer> v1, List<Integer> v2) {
        its = new Iterator[]{v1.iterator(), v2.iterator()};
        pos = 0;
    }

    public int next() {
        int next = its[pos].next();
        pos = (pos == its.length - 1) ? 0 : pos + 1;
        return next;
    }

    public boolean hasNext() {
        if (its[pos].hasNext()) return true;
        for (int i = pos + 1; i < its.length; i++) {
            if (its[i].hasNext()) {
                pos = i;
                return true;
            }
        }
        for (int i = 0; i < pos; i++) {
            if (its[i].hasNext()) {
                pos = i;
                return true;
            }
        }
        return false;
    }
}