solution.md

# 螺旋矩阵 II

<p>给你一个正整数 <code>n</code> ，生成一个包含 <code>1</code> 到 <code>n<sup>2</sup></code> 所有元素，且元素按顺时针顺序螺旋排列的 <code>n x n</code> 正方形矩阵 <code>matrix</code> 。</p><p> </p><p><strong>示例 1：</strong></p><img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0059.Spiral%20Matrix%20II/images/spiraln.jpg" style="width: 242px; height: 242px;" /><pre><strong>输入：</strong>n = 3<strong><br />输出：</strong>[[1,2,3],[8,9,4],[7,6,5]]</pre><p><strong>示例 2：</strong></p><pre><strong>输入：</strong>n = 1<strong><br />输出：</strong>[[1]]</pre><p> </p><p><strong>提示：</strong></p><ul>	<li><code>1 <= n <= 20</code></li></ul>

## template

```cpp
#include <bits/stdc++.h>
using namespace std;
class Solution
{
public:
	vector<vector<int>> generateMatrix(int n)
	{
		vector<vector<int>> matrix(n, vector<int>(n));
		int direction = 0;
		int hor_top = 0;
		int hor_bottom = n - 1;
		int ver_left = 0;
		int ver_right = n - 1;
		int num = 0;
		while (num < n * n)
		{
			switch (direction)
			{
			case 0:
				for (int i = ver_left; i <= ver_right; i++)
				{
					matrix[hor_top][i] = ++num;
				}
				hor_top++;
				break;
			case 1:
				for (int i = hor_top; i <= hor_bottom; i++)
				{
					matrix[i][ver_right] = ++num;
				}
				ver_right--;
				break;
			case 2:
				for (int i = ver_right; i >= ver_left; i--)
				{
					matrix[hor_bottom][i] = ++num;
				}
				hor_bottom--;
				break;
			case 3:
				for (int i = hor_bottom; i >= hor_top; i--)
				{
					matrix[i][ver_left] = ++num;
				}
				ver_left++;
				break;
			}
			direction++;
			direction %= 4;
		}
		return matrix;
	}
};
```

## 答案

```cpp

```

## 选项

### A

```cpp

```

### B

```cpp

```

### C

```cpp

```