Problem

Given an m x n matrix, return all elements of the matrix in spiral order.

https://leetcode.com/problems/spiral-matrix/

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

Test Cases

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class Solution:
def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
solution_test.py
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import pytest

from solution import Solution


@pytest.mark.parametrize('matrix, expected', [
([[1,2,3],[4,5,6],[7,8,9]], [1,2,3,6,9,8,7,4,5]),
([[1,2,3,4],[5,6,7,8],[9,10,11,12]], [1,2,3,4,8,12,11,10,9,5,6,7]),
])
class Test:
def test_solution(self, matrix, expected):
sol = Solution()
assert sol.spiralOrder(matrix) == expected

Thoughts

主要就是注意控制好边界。

Code

solution.py
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from typing import List


class Solution:
def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
m = len(matrix)
n = len(matrix[0])
res = [0] * (m * n)

steps = ((0, 1), (1, 0), (0, -1), (-1, 0)) # →, ↓, ←, ↑
limits = [n, m - 1] # ↔︎, ↕
i = 0
j = -1
dir = 0
move = 0
for idx in range(m * n):
i += steps[dir][0]
j += steps[dir][1]
res[idx] = matrix[i][j]
move += 1
if move == limits[dir & 1]:
limits[dir & 1] -= 1
dir = (dir + 1) % 4
move = 0

return res