Flatten 2D Vector

此题只需要两个指针就能搞定了，但是实际上的思路还是lazy iteration，也就是需要的时候我才往前走，不要的时候就赖在原地不停给结果，其实这也是所有迭代器类型的题目的思路，能不先遍历就尽量先不遍历，一旦遍历了就一次性多存储一些数，这道题还需要注意空列表的情况

class Vector2D(object):

    def __init__(self, vec2d):
        """
        Initialize your data structure here.
        :type vec2d: List[List[int]]
        """
        self.n = len(vec2d)
        self.vector = vec2d
        self.curRow = 0
        self.curCol = 0

    def next(self):
        """
        :rtype: int
        """
        result = self.vector[self.curRow][self.curCol]
        self.curCol += 1
        return result

    def hasNext(self):
        """
        :rtype: bool
        """
        while self.curRow < self.n and self.curCol >= len(self.vector[self.curRow]):
            self.curRow += 1
            self.curCol = 0
        return self.curRow < self.n


# Your Vector2D object will be instantiated and called as such:
# i, v = Vector2D(vec2d), []
# while i.hasNext(): v.append(i.next())