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?

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].

class ZigzagIterator(object):

    def __init__(self, v1, v2):
        """
        Initialize your data structure here.
        :type v1: List[int]
        :type v2: List[int]
        """
        self.v = [v1,v2]
        self.start = 0
        self.cur = 0
        self.len = len(v1) + len(v2)

    def next(self):
        """
        :rtype: int
        """
        while True:
            if self.v[self.cur%2]:
                temp = self.v[self.cur % 2].pop(0)
                self.start += 1
                self.cur += 1
                return temp
            else:
                self.cur += 1

    def hasNext(self):
        """
        :rtype: bool
        """
        if self.start < self.len:
            return True
        else:
            return False

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();
        int result = (Integer)poll.next();
        if(poll.hasNext()) list.add(poll);
        return result;
    }

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