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]第一种思路：
模拟法，直接模拟螺旋的顺序，用一个变量state 来作为控制方向的依据。
每次走过的点，就置为-999999标记来过了。
class Solution(object):
def spiralOrder(self, matrix):
"""
:type matrix: List[List[int]]
:rtype: List[int]
"""
m = len(matrix)
if m == 0:
return []
n = len(matrix[0])
if n == 0:
return []
x, y = 0,0
cnt = 1
state = "right"
res = list()
while(cnt <= m * n):
# print x, y,state
res.append(matrix[x][y])
matrix[x][y] = -999999 #标记来过
cnt += 1
if state == "right":
if y != n - 1 and matrix[x][y + 1] != -999999: #可以向右走
y += 1
continue
else:#要向下转了
if x + 1 < m and matrix[x + 1][y] != -999999: #可以向下
x += 1
state = "down"
continue
elif state == "down":
if x != m - 1 and matrix[x + 1][y] != -999999: #可以向下走
x += 1
continue
else:#要向左转
if y > 0 and matrix[x][y - 1] != -999999:
y -= 1
state = "left"
continue
elif state == "left":
if y != 0 and matrix[x][y - 1] != -999999:
y -= 1
continue
else:#要向上走
if x != 0 and matrix[x - 1][y] != -999999:
x -= 1
state = "up"
continue
elif state == "up":
if x != 0 and matrix[x - 1][y] != -999999:
x -= 1
continue
else:#右转
if y != n -1 and matrix[x][y + 1] != -999999:
y += 1
state = "right"
continue
# print res
return res
第二种思路：
还是模拟法，但是用方向变量的改变值di， dj来控制方向。
class Solution(object):
def spiralOrder(self, matrix):
"""
:type matrix: List[List[int]]
:rtype: List[int]
"""
m = len(matrix)
if m == 0:
return []
n = len(matrix[0])
if n == 0:
return []
r, i, j, di, dj = list(), 0, 0, 0, 1
for _ in range(m * n):
r.append(matrix[i][j])
matrix[i][j] = None
if matrix[(i + di) % m][(j + dj) % n] == None: #来过了改转向了
di, dj = dj, -di #0 1 变 1 0, 1 0 变 0 -1, 0 -1 变 -1, 0, -1 0 变 0 1
i += di
j += dj
return r