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60 lines
1.4 KiB
Plaintext
60 lines
1.4 KiB
Plaintext
---
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title: '54. Spiral Matrix'
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description: Given an m x n matrix, return all elements of the matrix in spiral order
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sidebar:
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label: 'Spiral Matrix'
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badge: 'Medium'
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---
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<Badge variant="accent">Matrix Index Math</Badge>
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### Example 1:
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- Input: `matrix = [[1,2,3],[4,5,6],[7,8,9]]`
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- Output: `[1,2,3,6,9,8,7,4,5]`
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### Example 2:
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- Input: `matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]`
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- Output: `[1,2,3,4,8,12,11,10,9,5,6,7]`
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### Constraints:
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- `m == matrix.length`
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- `n == matrix[i].length`
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- `1 <= m, n <= 10`
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- `-100 <= matrix[i][j] <= 100`
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## Solution
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```py
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class Solution:
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def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
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res = []
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top, bottom = 0, len(matrix) - 1
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left, right = 0, len(matrix[0]) - 1
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while top <= bottom and left <= right:
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# 1. Top Row
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for c in range(left, right + 1):
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res.append(matrix[top][c])
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top += 1
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# 2. Right Column
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for r in range(top, bottom + 1):
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res.append(matrix[r][right])
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right -= 1
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# 3. Bottom Row
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if top <= bottom:
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for c in range(right, left - 1, -1):
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res.append(matrix[bottom][c])
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bottom -= 1
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# 4. Left Column
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if left <= right:
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for r in range(bottom, top - 1, -1):
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res.append(matrix[r][left])
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left += 1
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return res
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```
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