54. Spiral Matrix

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

python
class Solution:
    def spiralOrder(self, matrix: List[List[int]]) -> List[int]:

solution_test.py下载完整源码
py
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下载完整源码
py
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

#Problem

#Test Cases

#Thoughts

#Code

Problem

Test Cases

Thoughts

Code