Given an m x n matrix, return all elements of the matrix in spiral order.
Example 1:
1 -> 2 -> 3
v
4 -> 5 6
^ v
7 <- 8 <- 9
Input: matrix = [[1,2,3],[4,5,6],[7,8,9]]
Output: [1,2,3,6,9,8,7,4,5]
Example 2:
1 -> 2 -> 3 -> 4
v
5 -> 6 -> 7 8
^ v
9 <- 10 < 11 < 12
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.lengthn == matrix[i].length1 <= m, n <= 10-100 <= matrix[i][j] <= 100
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class Solution:
def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
result = []
rows, columns = len(matrix), len(matrix[0])
up = left = 0
right = columns - 1
down = rows - 1
while len(result) < rows * columns:
# Traverse from left to right.
for col in range(left, right + 1):
result.append(matrix[up][col])
# Traverse downwards.
for row in range(up + 1, down + 1):
result.append(matrix[row][right])
if up != down:
# Traverse from right to left.
for col in range(right - 1, left - 1, -1):
result.append(matrix[down][col])
if left != right:
# Traverse upwards.
for row in range(down - 1, up, -1):
result.append(matrix[row][left])
left += 1
right -= 1
up += 1
down -= 1
return result
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