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]
class Solution:
def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
if not matrix:return []
res=[]
h_start=0
h_end=len(matrix)-1
l_start=0
l_end=len(matrix[0])-1
while h_start<=h_end and l_start<=l_end:
for i in range(l_start,l_end+1):
res.append(matrix[h_start][i])
h_start+=1
if h_start>h_end:break*
for i in range(h_start,h_end+1):
res.append(matrix[i][l_end])
l_end-=1
if l_start>l_end:break*
for i in range(l_end,l_start-1,-1):
res.append(matrix[h_end][i])
h_end-=1
if h_start>h_end:break*
for i in range(h_end,h_start-1,-1):
res.append(matrix[i][l_start])
l_start+=1
#print(l_start,l_end,h_start,h_end)
return res