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].



# Solution

class ZigzagIterator(object):

def __init__(self, v1, v2):
"""
:type v1: List[int]
:type v2: List[int]
"""
self.lists = list()
if len(v1) > 0:
self.lists.append(v1)
if len(v2) > 0:
self.lists.append(v2)

self.numLists = len(self.lists)
self.counters = [0] * self.numLists
self.listCounter = 0

def next(self):
"""
:rtype: int
"""
res = self.lists[self.listCounter][self.counters[self.listCounter]]
if self.counters[self.listCounter] == len(self.lists[self.listCounter]) - 1:
del self.counters[self.listCounter]
del self.lists[self.listCounter]
self.numLists -= 1
else:
self.counters[self.listCounter] += 1
self.listCounter += 1
if self.listCounter == self.numLists:
self.listCounter = 0
return res

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

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