> For the complete documentation index, see [llms.txt](https://kongchengzhuge.gitbook.io/leetcode/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://kongchengzhuge.gitbook.io/leetcode/48.-rotate-image.md).

# 48. Rotate Image

## problem description

You are given an *n* x *n* 2D matrix representing an image.

Rotate the image by 90 degrees (clockwise).

**Note:**

You have to rotate the image [**in-place**](https://en.wikipedia.org/wiki/In-place_algorithm), which means you have to modify the input 2D matrix directly. **DO NOT** allocate another 2D matrix and do the rotation.

**Example 1:**

```
Given input matrix = 
[
  [1,2,3],
  [4,5,6],
  [7,8,9]
],

rotate the input matrix in-place such that it becomes:
[
  [7,4,1],
  [8,5,2],
  [9,6,3]
]
```

**Example 2:**

```
Given input matrix =
[
  [ 5, 1, 9,11],
  [ 2, 4, 8,10],
  [13, 3, 6, 7],
  [15,14,12,16]
], 

rotate the input matrix in-place such that it becomes:
[
  [15,13, 2, 5],
  [14, 3, 4, 1],
  [12, 6, 8, 9],
  [16, 7,10,11]
]
```

## algorithm thought

最简单的方法肯定是定义一个矩阵，然后将这个矩阵转置之后赋值给这个矩阵。但是这样需要O(n²)空间题目也明显要求了我们空间复杂度维持在O(1)内。我们必须在本来的数组上操作。如果只能这么操作的话，肯定就是使用swap函数，交换值。我们可以对称的改变矩阵，对于旋转90度，我们只需要首先按照竖直中线对称，再按照对角线对称就能得到结果。

## code

```cpp
class Solution {
public:
    void rotate(vector<vector<int>>& matrix) {
        int n=matrix.size();
        for(int i=0;i<n/2;i++)
            for(int j=0;j<n;++j){
                swap(matrix[i][j],matrix[n-1-i][j]);
            }
        for(int i=0;i<n;i++)
            for(int j=i+1;j<n;++j){
                swap(matrix[i][j],matrix[j][i]);
            }
    }
};
```

## algorithm analysis

对二维矩阵操作，双重循环。最后时间复杂度是O(n²)