# Spiral Matrix

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

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:

```class Solution:
def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
ROWS, COLS = len(matrix), len(matrix[0])
top, left = 0, 0
right, bottom = COLS, ROWS
res = []
while left < right and top < bottom:
# get every i in top row
for i in range(left, right):
res.append(matrix[top][i])
top += 1
# get every i in right column
for i in range(top, bottom):
res.append(matrix[i][right - 1])
right -= 1

if not (top < bottom and left < right):
break
# get every i in bottom row
for i in range(right - 1, left - 1, -1):
res.append(matrix[bottom - 1][i])
bottom -= 1
# get every i in left col
for i in range(bottom - 1, top - 1, -1):
res.append(matrix[i][left])
left += 1

return res
```

Ashwini Verma on Jan 31, 2022 at 08:01 am

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