281. 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].

Solution:

class ZigzagIterator(object):
    def __init__(self, v1, v2):
        """
        Initialize your data structure here.
        :type v1: List[int]
        :type v2: List[int]
        """
        max_len = 0
        for v in v1, v2:
            max_len = max(max_len, len(v))

        self.nums = []
        for idx in xrange(max_len):
            for v in v1, v2:
                if idx < len(v):
                    self.nums.append(v[idx])
        self.nums.reverse()

    def next(self):
        """
        :rtype: int
        """
        return self.nums.pop()

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

Lessons:

• Pre-compute a list of all numbers in constructor.
• Use the length of longest vector.