给定 matrix = [[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]

# 列步进方向
# 顶，右，底，左
cds=[1, 0, -1, 0]# 行步进方向
# 顶，右，底，左
rds=[0, 1, 0, -1]

def spiralOrder(matrix):if not matrix:return []cds = [1, 0, -1, 0] # 列坐标步进方向rds = [0, 1, 0, -1] # 行坐标步进方向# 行列当前元素数量（步进数）rs, cs = len(matrix), len(matrix[0])turns = 0 # 拐弯次数r, c = 0, -1 # 当前元素坐标elem_num = rs * csresult = [0] * elem_num # 结果index = 0 # 当前结果下标while index < elem_num:di = turns % 4 # 步进方向下标# 每次拐弯，交替使用行和列的步进steps = cs if turns % 2 == 0 else rsfor i in range(steps):r += rds[di]c += cds[di]result[index] = matrix[r][c]index += 1# 每两次拐弯，行列的步进数递减if turns % 2 == 0:cs -= 1rs -= 1turns += 1return result

cd, rd = -rd, cd

def spiralOrder(matrix):if not matrix:return []cd, rd = 1, 0 # 列和行步进方向rs, cs = len(matrix), len(matrix[0])turns = 0r, c = 0, -1elem_num = rs * csresult = [0] * elem_numindex = 0while index < elem_num:if turns % 2 == 0:steps = cscs -= 1rs -= 1else:steps = rsfor i in range(steps):r += rdc += cdresult[index] = matrix[r][c]index += 1turns += 1# 每次拐弯，重新决定行列步进方向cd, rd = -rd, cdreturn result

def spiralOrder(matrix):if not matrix:return []cd, rd = 1, 0 # 列和行的步进方向rows, cols = len(matrix), len(matrix[0])# 用于记录每个元素的访问状态visited = [[False] * cols for _ in range(rows)]r, c = 0, 0elem_num = rows * colsresult = [0] * elem_numfor i in range(elem_num):result[i] = matrix[r][c]visited[r][c] = True# 下一步的坐标nr和nc# 如果不越界，或者未被访问，则下个坐标就是nr和nc# 否则拐弯nr, nc = r + rd, c + cdif 0 <= nr < rows and 0 <= nc < cols and not visited[nr][nc]:r, c = nr, ncelse:cd, rd = -rd, cd # 重算拐弯后的步进方向r, c = r + rd, c + cdreturn result