54. 螺旋矩阵

题目：

给定一个包含 m x n 个元素的矩阵（m 行, n 列），请按照顺时针螺旋顺序，返回矩阵中的所有元素。

示例 1:

输入:
[
 [ 1, 2, 3 ],
 [ 4, 5, 6 ],
 [ 7, 8, 9 ]
]
输出: [1,2,3,6,9,8,7,4,5]

示例 2:

输入:
[
  [1, 2, 3, 4],
  [5, 6, 7, 8],
  [9,10,11,12]
]
输出: [1,2,3,4,8,12,11,10,9,5,6,7]

别人的trick，递归+zip函数 

class Solution:
    def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
        return matrix and list(matrix.pop(0)) + self.spiralOrder(list(zip(*matrix))[::-1])

matrix = [[1,2,3],[4,5,6],[7,8,9]]  --> [1,2,3,6,9,8,7,4,5]
-------------------------------------------------------------------------------
┊ list(matrix.pop(0))=[1,2,3]   list(zip(*matrix))=[(4,7),(5,8),(6,9)][::-1]
┊    matrix = [(6,9),(5,8),(4,7)]
┊   ---------------------------------------------------------------------------
┊   ┊ list(matrix.pop(0))=[6,9]   list(zip(*matrix))=[(5,4),(8,7)][::-1]    
┊   ┊    matrix = [(8,7),(5,4)]
┊   ┊   -----------------------------------------------------------------------
┊   ┊   ┊ list(matrix.pop(0))=[8,7]   list(zip(*matrix))=[(5,),(4,)][::-1]
┊   ┊   ┊    matrix = [(4,),(5,)]
┊   ┊   ┊   -------------------------------------------------------------------
┊   ┊   ┊   ┊ list(matrix).pop(0))=[4]    list(zip(*matrix))=[(5,)][::-1]
┊   ┊   ┊   ┊    matrix = [(5,)]
┊   ┊   ┊   ┊   ---------------------------------------------------------------
┊   ┊   ┊   ┊   ┊ list(matrix).pop(0))=[5]    list(zip(*matrix))=[]
┊   ┊   ┊   ┊   ┊   matrix = []
┊   ┊   ┊   ┊   ┊   -----------------------------------------------------------
┊   ┊   ┊   ┊   ┊   ┊ [] and self.spiralOrder(list(zip(*matrix))[::-1])
┊   ┊   ┊   ┊   ┊   ┊  Returns the first null value without subsequent operations
-------------------------------------------------------------------------------