# 29.顺时针打印矩阵

## 题目描述

输入一个矩阵，按照从外向里以顺时针的顺序依次打印出每一个数字。

示例 1：

```cpp
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：

```cpp
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]
```

限制：

```cpp
0 <= matrix.length <= 100
0 <= matrix[i].length <= 100
```

注意：本题与主站 54 题相同：[https://leetcode-cn.com/problems/spiral-matrix/](https://leetcode-cn.com/problems/spiral-matrix/)

来源：力扣（LeetCode）

链接：[https://leetcode-cn.com/problems/shun-shi-zhen-da-yin-ju-zhen-lcof](https://leetcode-cn.com/problems/shun-shi-zhen-da-yin-ju-zhen-lcof)

[题解链接](https://leetcode-cn.com/problems/shun-shi-zhen-da-yin-ju-zhen-lcof/solution/swift-fei-chang-xing-xiang-hao-li-jie-de-vbhy/)

[GitHub](https://github.com/RyukieSama/RyukiePlayground/blob/main/%E5%89%91%E6%8C%87Offer.playground/Pages/29.%20%E9%A1%BA%E6%97%B6%E9%92%88%E6%89%93%E5%8D%B0%E7%9F%A9%E9%98%B5.xcplaygroundpage/Contents.swift)

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## 解题思路

*   出现需要变更方向的点的拐向是一定的：

    * 当前向➡️扫描，拐点一定向↓
    * 当前向↓扫描，拐点一定向←
    * 当前向←扫描，拐点一定向↑
    * 当前向↑扫描，拐点一定向➡️

    `flags`为标记数组，用来标识已经扫描过的下标

    在一个方向上：`扫描到边界` 或者 `扫描到已标记的坐标` 则触发转向

## 代码

```swift
class Solution {
    enum Dirction {
        /// x + 1
        case Right
        /// y + 1
        case Down
        /// x - 1
        case Left
        /// y - 1
        case Up
    }

    func spiralOrder(_ matrix: [[Int]]) -> [Int] {
        var result: [Int] = []
        let height = matrix.count

        guard let first = matrix.first, height > 0 else {
            return result
        }

        let width = first.count
        var x = 0, y = 0
        var dir: Dirction = .Right

        var flags: [[Bool]] = Array(repeating: Array(repeating: false, count: width), count: height)

        while x >= 0, y >= 0, x < width, y < height, result.count < width * height {
            flags[y][x] = true
            result.append(matrix[y][x])

            switch dir {
            case .Right:
                if x + 1 < width, flags[y][x + 1] == false {
                    x += 1
                }
                else {
                    dir = .Down
                    y += 1
                }
            case .Down:
                if y + 1 < height, flags[y + 1][x] == false {
                    y += 1
                }
                else {
                    dir = .Left
                    x -= 1
                }
            case .Left:
                if x - 1 >= 0, x - 1 < width, flags[y][x - 1] == false {
                    x -= 1
                }
                else {
                    dir = .Up
                    y -= 1
                }
            case .Up:
                if y - 1 >= 0, y - 1 < height, flags[y - 1][x] == false {
                    y -= 1
                }
                else {
                    dir = .Right
                    x += 1
                }
            }
        }

        return result
    }
}
```
