# 螺旋矩阵
给你一个 m 行 n 列的矩阵 matrix ，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：matrix = [[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：matrix = [[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]
提示：
m == matrix.lengthn == matrix[i].length1 <= m, n <= 10-100 <= matrix[i][j] <= 100
## template
```java
class Solution {
public List spiralOrder(int[][] matrix) {
List res = new ArrayList();
if (matrix.length == 0 || (matrix.length == 1 && matrix[0].length == 0))
return res;
int left = 0;
int right = matrix[0].length - 1;
int top = 0;
int bottom = matrix.length - 1;
int num = (right + 1) * (bottom + 1);
while (num > 0) {
for (int j = left; j <= right; j++) {
res.add(matrix[top][j]);
num--;
}
if (num <= 0)
break;
top++;
for (int i = top; i <= bottom; i++) {
res.add(matrix[i][right]);
num--;
}
if (num <= 0)
break;
right--;
for (int j = right; j >= left; j--) {
res.add(matrix[bottom][j]);
num--;
}
if (num <= 0)
break;
bottom--;
for (int i = bottom; i >= top; i--) {
res.add(matrix[i][left]);
num--;
}
if (num <= 0)
break;
left++;
}
return res;
}
}
```
## 答案
```java
```
## 选项
### A
```java
```
### B
```java
```
### C
```java
```