Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.
Example 1:
Input:
[
[ 1, 2, 3 ],
[ 4, 5, 6 ],
[ 7, 8, 9 ]
]
Output: [1,2,3,6,9,8,7,4,5]
Example 2:
Input:
[
[1, 2, 3, 4],
[5, 6, 7, 8],
[9,10,11,12]
]
Output: [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 []
i = j = 0
m = len(matrix)
n = len(matrix[0])
marked = [[False] * n for _ in matrix]
step = [[0, 1], [1, 0], [0, -1], [-1, 0]]
i = j = di = 0
ans = []
for _ in range(m*n):
ans.append(matrix[i][j])
marked[i][j] = True
ni, nj = i + step[di][0], j + step[di][1]
if 0 <= ni < m and 0 <= nj < n and not marked[ni][nj]:
i, j = ni, nj
else:
di = (di + 1) % 4
i, j = i+step[di][0], j + step[di][1]
return ans