Leetcode: Zigzag Iterator

Posted on August 26, 2019November 16, 2019 by braindenny

Zigzag Iterator

Similar Problems:

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].

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

Solution: Use queue

## https://code.dennyzhang.com/zigzag-iterator
## Basic Ideas: array
##
## Complexity: Time O(1), Space O(1)
class ZigzagIterator(object):
    def __init__(self, v1, v2):
        """
        Initialize your data structure here.
        :type v1: List[int]
        :type v2: List[int]
        """
        self.queue = collections.deque([])
        if len(v1) != 0:
            self.queue.append((0, v1))
        if len(v2) != 0:
            self.queue.append((0, v2))

    def next(self):
        """
        :rtype: int
        """
        (i, v) = self.queue.popleft()
        res = v[i]
        i += 1
        if i != len(v):
            self.queue.append((i, v))
        return res

    def hasNext(self):
        """
        :rtype: bool
        """
        return len(self.queue) != 0

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

Solution: Use list

## https://code.dennyzhang.com/zigzag-iterator
## Basic Ideas: array
##
## Complexity: Time O(1), Space O(1)
class ZigzagIterator(object):
    def __init__(self, v1, v2):
        """
        Initialize your data structure here.
        :type v1: List[int]
        :type v2: List[int]
        """
        self.l = [v1, v2]
        self.indices = [0, 0]
        self.pos = 0

    def next(self):
        """
        :rtype: int
        """
        # skip finished list
        if self.indices[self.pos] == len(self.l[self.pos]):
            self.pos = (self.pos + 1)%len(self.l)

        index = self.indices[self.pos]
        res = self.l[self.pos][index]
        self.indices[self.pos] += 1
        # flip pointer
        self.pos = (self.pos + 1)%len(self.l)

        # TODO: how to avoid the overhead of finished list
        return res

    def hasNext(self):
        """
        :rtype: bool
        """
        for i in range(len(self.l)):
            if self.indices[i] != len(self.l[i]):
                return True
        return False

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

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