281. Zigzag Iterator

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].

解题要点：

用一个全局变量记录两个list加起来的长度。每进一次next，长度减去一，然后在while loop里的两个list里pop出来一个数，返回这个数。（或者可以用linkedlist和iterator，可解决k个数的问题）

class ZigzagIterator(object):

    def __init__(self, v1, v2):
        """
        Initialize your data structure here.
        :type v1: List[int]
        :type v2: List[int]
        """
        self.isV1 = True
        self.v1 = v1
        self.v2 = v2
        self.size = len(v1) + len(v2)

    def next(self):
        """
        :rtype: int
        """
        self.size -= 1
        res = 0
        while self.size != len(self.v1) + len(self.v2):
            if self.isV1 and len(self.v1) != 0:
                res = self.v1.pop(0)
                self.isV1 = False
            elif len(self.v2) != 0:
                res = self.v2.pop(0)
                self.isV1 = True
            else:
                self.isV1 = not self.isV1
        return res

    def hasNext(self):
        """
        :rtype: bool
        """
        return self.size > 0

# Your ZigzagIterator object will be instantiated and called as such:
# i, v = ZigzagIterator(v1, v2), []
# while i.hasNext(): v.append(i.next())