# 螺旋矩阵

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> [螺旋矩阵\_牛客题霸\_牛客网](https://www.nowcoder.com/practice/7edf70f2d29c4b599693dc3aaeea1d31)

**问题简述**

```
给定一个m x n大小的矩阵（m行，n列），按螺旋的顺序返回矩阵中的所有元素。
```

**思路**

* 把矩阵及其元素看作坐标中的点，初始定义左上角和右下角两个点 `(0, 0)` 和 `(m, n)`；
* 然后根据要求，每次打印一圈，注意边界条件，详见代码；

<details>

<summary><strong>Python</strong></summary>

```python
#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param matrix int整型二维数组 
# @return int整型一维数组
#
class Solution:
    def spiralOrder(self , M: List[List[int]]) -> List[int]:
        # write code here
        if not M: return []
        
        ret = []
        
        def dfs(a, b, c, d):
            
            # 打印一行 (a,b) -> (a, d)
            if a == c:
                t = b
                while t <= d:
                    ret.append(M[a][t])
                    t += 1
            # 打印一列 (a, d) -> (c, d)
            elif b == d:
                t = a
                while t <= c:
                    ret.append(M[t][b])
                    t += 1
            # 打印一圈
            else:
                # 左到右: (a,b) -> (a, d-1)
                t = b
                while t < d:
                    ret.append(M[a][t])
                    t += 1
                # 上到下: (a, d) -> (c-1, d)
                t = a
                while t < c:
                    ret.append(M[t][d])
                    t += 1
                # 右到左: (c, d) -> (c, b+1)
                t = d
                while t > b:
                    ret.append(M[c][t])
                    t -= 1
                # 下到上: (c, b) -> (a-1, b)
                t = c
                while t > a:
                    ret.append(M[t][b])
                    t -= 1
        
        a, b, c, d = 0, 0, len(M) - 1, len(M[0]) - 1
        while a <= c and b <= d:
            dfs(a, b, c, d)
            a, b = a + 1, b + 1
            c, d = c - 1, d - 1
        
        return ret
```

</details>