class Solution:def spiralOrder(self, matrix: List[List[int]]) -> List[int]:# 用四个数对应4个遍历的方向[0, 1, 2, 3] - [右，下，左，上]go_state = 0 # 起始必须向右# record_matrix = [[0] * n for _ in range(m)]n_0, n_1, n_2, n_3 = 0, 0, 0, 0m, n = len(matrix), len(matrix[0]) go_num, limit_num = 0, len(matrix) * len(matrix[0])ans = []i, j = 0, 0while go_num < limit_num :ans.append(matrix[i][j])# record_matrix[i][j] = 1go_num += 1if go_state == 0:if j == n - 1 - n_1:go_state = 1i += 1n_0 += 1else:j += 1elif go_state == 1:if i == m - 1 - n_2:go_state = 2 j -= 1n_1 += 1else:i += 1elif go_state == 2:if j == n_3:go_state = 3 i -= 1n_2 += 1else:j -= 1else:if i == n_1:go_state = 0 j += 1n_3 += 1else:i -= 1return ans个人解法：
1.用4个状态标记遍历的走向
2.当前状态的结束状态依赖于下一个状态的完全执行次数
3.用遍历的元素数量作为退出循环的临界
大佬们的解法：
这种解法和大佬的方法二思路一致，其方法一则将当前状态以及该状态下对当前坐标的动作融合起来了，可以参考：https://leetcode.cn/problems/spiral-matrix/solutions/2966229/liang-chong-fang-fa-jian-ji-gao-xiao-pyt-4wzk
