Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.
Examples:
Input:
[
[ 1, 2, 3 ],
[ 4, 5, 6 ],
[ 7, 8, 9 ]
]
Output: [1,2,3,6,9,8,7,4,5]
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(object):
def spiralOrder(self, matrix):
"""
:type matrix: List[List[int]]
:rtype: List[int]
"""
if not matrix: return []
direction = 0
left = 0
right = len(matrix[0])-1
up = 0
down = len(matrix)-1
res = []
while left <= right and up <= down:
if direction == 0: # from left to right
for j in range(left, right+1):
res.append(matrix[up][j])
up += 1
if direction == 1: # from up to bottom
for i in range(up, down+1):
res.append(matrix[i][right])
right -= 1
if direction == 2: # from right to left
for j in range(right, left-1, -1):
res.append(matrix[down][j])
down -= 1
if direction == 3: # from bottom to up
for i in range(down, up-1, -1):
res.append(matrix[i][left])
left += 1
direction = (direction+1) % 4
return res