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# \[54]\[中等] 螺旋矩阵

## 题目描述

[54. 螺旋矩阵](https://leetcode-cn.com/problems/spiral-matrix/) [剑指 Offer 29. 顺时针打印矩阵](https://leetcode-cn.com/problems/shun-shi-zhen-da-yin-ju-zhen-lcof/) [59. 螺旋矩阵 II](https://leetcode-cn.com/problems/spiral-matrix-ii/)

给定一个包含 m x n 个元素的矩阵（m 行, n 列），请按照顺时针螺旋顺序，返回矩阵中的所有元素。

示例 1:

```
输入:
[
 [ 1, 2, 3 ],
 [ 4, 5, 6 ],
 [ 7, 8, 9 ]
]
输出: [1,2,3,6,9,8,7,4,5]
```

示例 2:

```
输入:
[
  [1, 2, 3, 4],
  [5, 6, 7, 8],
  [9,10,11,12]
]
输出: [1,2,3,4,8,12,11,10,9,5,6,7]
```

## 解题思路

[Spiral Matrix II （模拟法，设定边界，代码简短清晰）](https://leetcode-cn.com/problems/spiral-matrix-ii/solution/spiral-matrix-ii-mo-ni-fa-she-ding-bian-jie-qing-x/)

上面这种思路, 边界清晰. 外层循环每次走一圈, 循环内部, 依次循环顶部, 右部, 底部, 左部, 整个过程判断是否收集了足够数量的元素, 收集到足够元素后停止.

```python
class Solution:
    def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
        if len(matrix) == 0:
            return []
        n, m = len(matrix), len(matrix[0])
        num_total, res = n * m, []
        left, right, top, bottom = 0, m - 1, 0, n - 1
        while len(res) < num_total:
            if len(res) < num_total:
                for j in range(left, right + 1):
                    res.append(matrix[top][j])
                top += 1

            if len(res) < num_total:
                for i in range(top, bottom + 1):
                    res.append(matrix[i][right])
                right -= 1

            if len(res) < num_total:
                for j in range(right, left - 1, -1):
                    res.append(matrix[bottom][j])
                bottom -= 1

            if len(res) < num_total:
                for i in range(bottom, top - 1, -1):
                    res.append(matrix[i][left])
                left += 1
        return res
```