算法：由外向内顺时针打印矩阵（n*n）元素

输入一个矩阵(n*n)，按照从外向里以顺时针的顺序依次打印出每一个数字。
示例 1：
输入：matrix = [[1,2,3],[4,5,6],[7,8,9]]
输出：[1,2,3,6,9,8,7,4,5]
限制：
0 <= matrix.length
0 <= matrix[i].length 

import copy

class Solution(object):

    def __init__(self):
        self.lis = []

    def spiralOrder(self, matrix):
        """
        :type matrix: List[List[int]]
        :rtype: List[int]
        """
        rows = rols = len(matrix)
        if rows < 1:
            return []
        if rows == 1:
            self.lis.append(matrix[0][0])
            return self.lis
        if rows == 2:
            self.lis.append(matrix[0][0])
            self.lis.append(matrix[0][1])
            self.lis.append(matrix[1][1])
            self.lis.append(matrix[1][0])
            return self.lis
        if rows == 3:
            for i in matrix[0]:
                self.lis.append(i)
            for i in range(1, rols-1):
                self.lis.append(matrix[i][rols-1])
            for i in range(rols-1, -1, -1):
                self.lis.append(matrix[rols-1][i])
            for i in range(rols-2, 0, -1):
                self.lis.append(matrix[i][0])
            self.lis.append(matrix[1][1])
            return self.lis
        if rows > 3:
            for i in matrix[0]:
                self.lis.append(i)
            for i in range(1, rols-1):
                self.lis.append(matrix[i][rols-1])
            for i in range(rols-1, -1, -1):
                self.lis.append(matrix[rols-1][i])
            for i in range(rols-2, 0, -1):
                self.lis.append(matrix[i][0])
            lic = copy.copy(matrix)
            lic.pop(0)
            lic.pop()
            for i in lic:
                i.pop(0)
                i.pop()
            self.spiralOrder(lic)
            return self.lis

s = Solution()

liy = []
lix = []
n = 3
for i in range(1, n*n+1):
    liy.append(i)
for i in range(n):
    lix.append(liy[i*n : i*n + n])


ret = s.spiralOrder(lix)
print(ret)