Return a Matrix's Elements in Clockwise Spiral Order

Given an `m x n` matrix of integers, return all of its elements in clockwise spiral order. Start at the top-left element and walk right along the top row, down the rightmost column, left along the bottom row and up the leftmost column; then repeat the same clockwise walk on the remaining inner rectangle, and so on, until every element has been visited exactly once. ### Function Signature ```python def spiral_order(matrix: list[list[int]]) -> list[int]: ``` ### Rules - Every row of `matrix` has the same length. - Each element appears exactly once in the output, so the output has `m * n` values. - When the remaining rectangle is a single row, it is read left to right; when it is a single column, it is read top to bottom. ### Constraints - `1 <= m, n <= 200`, where `m` is the number of rows and `n` the number of columns. - `-10^4 <= matrix[i][j] <= 10^4` ### Examples **Example 1** - Input: `matrix = [[2, 4, 6], [8, 10, 12], [14, 16, 18], [20, 22, 24]]` - Output: `[2, 4, 6, 12, 18, 24, 22, 20, 14, 8, 10, 16]` - Explanation: the outer ring gives `2, 4, 6` (top row), `12, 18, 24` (right column), `22, 20` (bottom row, right to left) and `14, 8` (left column, bottom to top). The remaining inner column `10, 16` is read top to bottom. **Example 2** - Input: `matrix = [[5], [-1], [3]]` - Output: `[5, -1, 3]` **Example 3** - Input: `matrix = [[1, 2, 3, 4], [5, 6, 7, 8]]` - Output: `[1, 2, 3, 4, 8, 7, 6, 5]`