# 54. Spiral Matrix (M)

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

 

**Example 1:**

![](https://assets.leetcode.com/uploads/2020/11/13/spiral1.jpg)

```
Input: matrix = [[1,2,3],[4,5,6],[7,8,9]]
Output: [1,2,3,6,9,8,7,4,5]
```

**Example 2:**

![](https://assets.leetcode.com/uploads/2020/11/13/spiral.jpg)

```
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:

**解题的核心思路是按照右、下、左、上的顺序遍历数组，并使用四个变量圈定未遍历元素的边界**：

[![](https://labuladong.github.io/algo/images/%E8%8A%B1%E5%BC%8F%E9%81%8D%E5%8E%86/6.png)](https://labuladong.github.io/algo/images/%E8%8A%B1%E5%BC%8F%E9%81%8D%E5%8E%86/6.png)

随着螺旋遍历，相应的边界会收缩，直到螺旋遍历完整个数组：

[![](https://labuladong.github.io/algo/images/%E8%8A%B1%E5%BC%8F%E9%81%8D%E5%8E%86/7.png)](https://labuladong.github.io/algo/images/%E8%8A%B1%E5%BC%8F%E9%81%8D%E5%8E%86/7.png)

只要有了这个思路，翻译出代码就很容易了：

```java
List<Integer> spiralOrder(int[][] matrix) {
    int m = matrix.length, n = matrix[0].length;
    int upper_bound = 0, lower_bound = m - 1;
    int left_bound = 0, right_bound = n - 1;
    List<Integer> res = new LinkedList<>();
    // res.size() == m * n 则遍历完整个数组
    while (res.size() < m * n) {
        if (upper_bound <= lower_bound) {
            // 在顶部从左向右遍历
            for (int j = left_bound; j <= right_bound; j++) {
                res.add(matrix[upper_bound][j]);
            }
            // 上边界下移
            upper_bound++;
        }
        
        if (left_bound <= right_bound) {
            // 在右侧从上向下遍历
            for (int i = upper_bound; i <= lower_bound; i++) {
                res.add(matrix[i][right_bound]);
            }
            // 右边界左移
            right_bound--;
        }
        
        if (upper_bound <= lower_bound) {
            // 在底部从右向左遍历
            for (int j = right_bound; j >= left_bound; j--) {
                res.add(matrix[lower_bound][j]);
            }
            // 下边界上移
            lower_bound--;
        }
        
        if (left_bound <= right_bound) {
            // 在左侧从下向上遍历
            for (int i = lower_bound; i >= upper_bound; i--) {
                res.add(matrix[i][left_bound]);
            }
            // 左边界右移
            left_bound++;
        }
    }
    return res;
}
```


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