边工作边刷题：70天一遍leetcode: day 85-3

Zigzag Iterator

要点：

• 实际不是zigzag而是纵向访问
• 这题可以扩展到k个list，也可以扩展到只给iterator而不给list。结构上没什么区别，iterator的hasNext和计数殊途同归。
• 一种方法利用queue，进queue的顺序就是下一个要访问的元素index和v，实际就是level order traversal
• 另一种是用一个list存每个v的当前位置（or iterator）。然后track count：当前列访问完了（即list访问一遍），reset count，而如果一列元素没了，就把当前的v从list中去掉。
• 两种方法其实都是level order traversal的方式。
• python的iter没有hasNext()而是throw exception，所以就不用iterator实现了。

https://repl.it/CfG1/1

# 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].
# Hide Company Tags Google
# Hide Tags Design
# Hide Similar Problems (M) Binary Search Tree Iterator (M) Flatten 2D Vector (M) Peeking Iterator (M) Flatten Nested List Iterator

from collections import deque
class ZigzagIterator(object):

    def __init__(self, v1, v2):
        """
        Initialize your data structure here.
        :type v1: List[int]
        :type v2: List[int]
        """
        self.data = deque([(0, v) for v in (v1, v2) if v])

    def next(self):
        """
        :rtype: int
        """
        l, v = self.data.popleft()
        ret = v[l]
        if l<len(v)-1:
        	self.data.append((l+1, v))
        return ret

    def hasNext(self):
        """
        :rtype: bool
        """
        return bool(self.data)

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