# 54.Spiral-Matrix

## 54. Spiral Matrix

## 题目地址

<https://leetcode.com/problems/spiral-matrix/>

## 题目描述

```
Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.

Example 1:

Input:
[
 [ 1, 2, 3 ],
 [ 4, 5, 6 ],
 [ 7, 8, 9 ]
]
Output: [1,2,3,6,9,8,7,4,5]
Example 2:

Input:
[
  [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]
```

## 代码

### Approach 1: Simulation

**Complexity Analysis**

* Time Complexity: &#x4F;*(\_N*), where N\_ is the total number of elements in the input matrix. We add every element in the matrix to our final answer.
* Space Complexity: O\_(\_N), the information stored in `seen` and in `ans`.

```java
class Solution {
  public List<Integer> spiralOrder(int[][] matrix) {
    List ans = new ArrayList();
    if (matrix.length == 0) return ans;
    int R = matrix.length;
    int C = matrix[0].length;

    boolean[][] seen = new boolean[R][C];
    int[] dr = {0, 1, 0, -1};
    int[] dc = {1, 0, -1, 0}; // top, right, bottom, left
         int r = 0, c = 0, di = 0;
    for (int i = 0; i < R * C; i++) {
      ans.add(matrix[r][c]);
      seen[r][c] = true;
      int cr = r + dr[di];
      int cc = c + dc[di];
      if (cr >= 0 && cr < R && cc >= 0 && cc < C && !seen[cr][cc]) {
        r = cr;
        c = cc;
      } else {
        di = (di + 1) % 4;
        r += dr[di];
        c += dc[di];
      }
    }

    return ans;
  }
}
```

### Approach 2: Layer-by-Layer

```java
class Solution {
  public List<Integer> spiralOrder(int[][] matrix) {
    List ans = new ArrayList();
    if (matrix.length == 0) return ans;
    int r1 = 0, r2 = matrix.length - 1;
    int c1 = 0, c2 = matrix.length[0] - 1;
    while (r1 <= r2 && c1 <= c2) {
      // top
      for (int c = c1; c <= c2; c++) {
        ans.add(matrix[r1][c]);
      }
      // right
      for (int r = r1 + 1; r <= r2; r++) {
        ans.add(matrix[r][c2]);
      }

      if (r1 < r2 && c1 < c2) { // 奇数最后一轮无法进行
        // bottom
        for (int c = c2 - 1; c > c1; c--) {
          ans.add(matrix[r2][c]);
        }
        // left
        for (int r = r2; r < r1; r--) {
          ans.add(matrix[r][c1]);
        }
      }

      r1++;
      r2--;
      c1++;
      c2--;
    }

    return ans;
  }
}
```


---

# Agent Instructions: Querying This Documentation

If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter:

```
GET https://wentao-shao.gitbook.io/leetcode/matrix/spiral-matrix/54.spiral-matrix.md?ask=<question>
```

The question should be specific, self-contained, and written in natural language.
The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.