algorithm-base/animation-simulation/数组篇/leetcode59螺旋矩阵2.md

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### [59.螺旋矩阵 II](https://leetcode-cn.com/problems/spiral-matrix-ii)

给你一个正整数 `n` ，生成一个包含 `1` 到 `n2` 所有元素，且元素按顺时针顺序螺旋排列的 `n x n` 正方形矩阵 `matrix` 。

**示例 1：**

> 输入：n = 3
> 输出：[[1,2,3],[8,9,4],[7,6,5]]

**示例 2：**

> 输入：n = 1
> 输出：[[1]]

其实我们只要做过了螺旋矩阵 第一题，这个题目我们完全可以一下搞定，几乎没有进行更改，我们先来看下 **leetcode 54** 题的解析。

### leetcode 54 螺旋矩阵

题目描述

_给定一个包含 m_ x n 个元素的矩阵（m 行, n 列），请按照顺时针螺旋顺序，返回矩阵中的所有元素。

示例一

> 输入：matrix = [[1,2,3],[4,5,6],[7,8,9]]
> 输出：[1,2,3,6,9,8,7,4,5]

示例二

> 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]

这个题目很细非常细，思路很容易想到，但是要是完全实现也不是特别容易，我们一起分析下这个题目，我们可以这样理解，我们像剥洋葱似的一步步的剥掉外皮，直到遍历结束，见下图。

_![螺旋矩阵](https://pic.leetcode-cn.com/1615813563-uUiWlF-file_1615813563382)_

题目很容易理解，但是要想完全执行出来，也是不容易的，因为这里面的细节太多了，我们需要认真仔细的考虑边界。

我们也要考虑重复遍历的情况即什么时候跳出循环。刚才我们通过箭头知道了我们元素的遍历顺序，这个题目也就完成了一大半了，下面我们来讨论一下什么时候跳出循环，见下图。

注：这里需要注意的是，框框代表的是每个边界。

![](https://img-blog.csdnimg.cn/20210318095839543.gif)

题目代码：

Java Code:

```java
class Solution {
    public List<Integer> spiralOrder(int[][] matrix) {

        List<Integer> arr = new ArrayList<>();
        int left = 0, right = matrix[0].length-1;
        int top = 0, down = matrix.length-1;

        while (true) {
             for (int i = left; i <= right; ++i) {
                 arr.add(matrix[top][i]);
             }
             top++;
             if (top > down) break;
             for (int i = top; i <= down; ++i) {
                 arr.add(matrix[i][right]);
             }
             right--;
             if (left > right) break;
             for (int i = right; i >= left; --i) {
                 arr.add(matrix[down][i]);
             }
             down--;
             if (top > down) break;
             for (int i = down; i >= top; --i) {
                 arr.add(matrix[i][left]);
             }
             left++;
             if (left > right) break;

        }
        return arr;
    }
}

```

Python3 Code:

```python
from typing import List
class Solution:
    def spiralOrder(self, matrix: List[List[int]])->List[int]:
        arr = []
        left = 0
        right = len(matrix[0]) - 1
        top = 0
        down = len(matrix) - 1
        while True:
            for i in range(left, right + 1):
                arr.append(matrix[top][i])
            top += 1
            if top > down:
                break
            for i in range(top, down + 1):
                arr.append(matrix[i][right])
            right -= 1
            if left > right:
                break
            for i in range(right, left - 1, -1):
                arr.append(matrix[down][i])
            down -= 1
            if top > down:
                break
            for i in range(down, top - 1, -1):
                arr.append(matrix[i][left])
            left += 1
            if left > right:
                break
        return arr
```

C++ Code:

```cpp
class Solution {
public:
    vector<int> spiralOrder(vector<vector<int>>& matrix) {
        vector <int> arr;
        int left = 0, right = matrix[0].size()-1;
        int top = 0, down = matrix.size()-1;
        while (true) {
             for (int i = left; i <= right; ++i) {
                 arr.emplace_back(matrix[top][i]);
             }
             top++;
             if (top > down) break;
             for (int i = top; i <= down; ++i) {
                 arr.emplace_back(matrix[i][right]);
             }
             right--;
             if (left > right) break;
             for (int i = right; i >= left; --i) {
                 arr.emplace_back(matrix[down][i]);
             }
             down--;
             if (top > down) break;
             for (int i = down; i >= top; --i) {
                 arr.emplace_back(matrix[i][left]);
             }
             left++;
             if (left > right) break;
        }
        return arr;
    }
};
```

Swift Code:

```swift
class Solution {
    func spiralOrder(_ matrix: [[Int]]) -> [Int] {
        var arr:[Int] = []
        var left = 0, right = matrix[0].count - 1
        var top = 0, down = matrix.count - 1

        while (true) {
            for i in left...right {
                arr.append(matrix[top][i])
            }
            top += 1
            if top > down { break }
            for i in top...down {
                arr.append(matrix[i][right])
            }
            right -= 1
            if left > right { break}
            for i in stride(from: right, through: left, by: -1) {
                arr.append(matrix[down][i])
            }
            down -= 1
            if top > down { break}
            for i in stride(from: down, through: top, by: -1) {
                arr.append(matrix[i][left])
            }
            left += 1
            if left > right { break}
        }

        return arr
    }
}
```

我们仅仅是将 54 反过来了，往螺旋矩阵里面插值，下面我们直接看代码吧,大家可以也可以对其改进，大家可以思考一下，如果修改能够让代码更简洁！

Java Code:

```java
class Solution {
    public int[][] generateMatrix(int n) {

        int[][] arr = new int[n][n];
        int left = 0;
        int right = n-1;
        int top = 0;
        int buttom = n-1;
        int num = 1;
        int numsize = n*n;
        while (true) {
            for (int i = left; i <= right; ++i) {
                arr[top][i] = num++;
            }
            top++;
            if (num > numsize) break;
            for (int i = top; i <= buttom; ++i) {
                arr[i][right] = num++;

            }
            right--;
            if (num > numsize) break;
            for (int i = right; i >= left; --i) {
                arr[buttom][i] = num++;
            }
            buttom--;
            if (num > numsize) break;
            for (int i = buttom; i >= top; --i) {
                arr[i][left] = num++;
            }
            left++;
            if (num > numsize) break;

        }
        return arr;
    }
}
```

Python3 Code:

```python
from typing import List
import numpy as np
class Solution:
    def generateMatrix(self, n: int)->List[List[int]]:
        arr = np.array([[0] * n] * n)
        left = 0
        right = n - 1
        top = 0
        buttom = n - 1
        num = 1
        numsize = n * n
        while True:
            for i in range(left, right + 1):
                arr[top][i] = num
                num += 1
            top += 1
            if num > numsize:
                break
            for i in range(top, buttom + 1):
                arr[i][right] = num
                num += 1
            right -= 1
            if num > numsize:
                break
            for i in range(right, left - 1, -1):
                arr[buttom][i] = num
                num += 1
            buttom -= 1
            if num > numsize:
                break
            for i in range(buttom, top - 1, -1):
                arr[i][left] = num
                num += 1
            left += 1
            if num > numsize:
                break
        return arr.tolist()
```

C++ Code:

```cpp
class Solution {
public:
    vector<vector<int>> generateMatrix(int n) {
        vector <vector <int>> arr(n, vector <int>(n));
        int left = 0, right = n-1, top = 0, buttom = n - 1, num = 1, numsize = n * n;
        while (true) {
            for (int i = left; i <= right; ++i) {
                arr[top][i] = num++;
            }
            top++;
            if (num > numsize) break;
            for (int i = top; i <= buttom; ++i) {
                arr[i][right] = num++;
            }
            right--;
            if (num > numsize) break;
            for (int i = right; i >= left; --i) {
                arr[buttom][i] = num++;
            }
            buttom--;
            if (num > numsize) break;
            for (int i = buttom; i >= top; --i) {
                arr[i][left] = num++;
            }
            left++;
            if (num > numsize) break;

        }
        return arr;
    }
};
```

Swift Code:

```swift
class Solution {
    func generateMatrix(_ n: Int) -> [[Int]] {
        var arr:[[Int]] = Array.init(repeating: Array.init(repeating: 0, count: n), count: n)
        var left = 0, right = n - 1
        var top = 0, bottom = n - 1
        var num = 1, numSize = n * n

        while true {
            for i in left...right {
                arr[top][i] = num
                num += 1
            }
            top += 1
            if num > numSize { break}
            for i in top...bottom {
                arr[i][right] = num
                num += 1
            }
            right -= 1
            if num > numSize { break}
            for i in stride(from: right, through: left, by: -1) {
                arr[bottom][i] = num
                num += 1
            }
            bottom -= 1
            if num > numSize { break}
            for i in stride(from: bottom, through: top, by: -1) {
                arr[i][left] = num
                num += 1
            }
            left += 1
            if num > numSize { break}
        }

        return arr
    }
}
```