Given an m x n matrix, return all elements of the matrix in spiral order.
Example 1:
Input: matrix = [[1,2,3],[4,5,6],[7,8,9]]
Output: [1,2,3,6,9,8,7,4,5]
Example 2:
Input: matrix = [[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]
Constraints:
-
m == matrix.length -
n == matrix[i].length -
1 <= m, n <= 10 -
-100 <= matrix[i][j] <= 100
SOLUTION:
class Solution:
def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
m = len(matrix)
n = len(matrix[0])
op = []
mark = "X"
ctr = 0
direction = 0
x, y = 0, 0
while ctr < m * n:
if matrix[x][y] != mark:
ctr += 1
op.append(matrix[x][y])
matrix[x][y] = mark
switch = False
if direction == 0:
if (y + 1) < n and matrix[x][y + 1] != mark:
y += 1
else:
switch = True
elif direction == 1:
if (x + 1) < m and matrix[x + 1][y] != mark:
x += 1
else:
switch = True
elif direction == 2:
if (y - 1) >= 0 and matrix[x][y - 1] != mark:
y -= 1
else:
switch = True
else:
if (x + 1) >= 0 and matrix[x - 1][y] != mark:
x -= 1
else:
switch = True
if switch:
direction = (direction + 1) % 4
return op
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