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---
title: '54. Spiral Matrix'
description: Given an m x n matrix, return all elements of the matrix in spiral order
sidebar:
label: 'Spiral Matrix'
badge: 'Medium'
---
<Badge variant="accent">Matrix Index Math</Badge>
### 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
```py
class Solution:
def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
res = []
top, bottom = 0, len(matrix) - 1
left, right = 0, len(matrix[0]) - 1
while top <= bottom and left <= right:
# 1. Top Row
for c in range(left, right + 1):
res.append(matrix[top][c])
top += 1
# 2. Right Column
for r in range(top, bottom + 1):
res.append(matrix[r][right])
right -= 1
# 3. Bottom Row
if top <= bottom:
for c in range(right, left - 1, -1):
res.append(matrix[bottom][c])
bottom -= 1
# 4. Left Column
if left <= right:
for r in range(bottom, top - 1, -1):
res.append(matrix[r][left])
left += 1
return res
```