# [48. Rotate Image](https://leetcode.com/problems/rotate-image/)

## 题目

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**:

![](https://assets.leetcode.com/uploads/2020/08/28/mat1.jpg)

    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**:

![](https://assets.leetcode.com/uploads/2020/08/28/mat2.jpg)

    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]
    ]


## 题目大意

给定一个 n × n 的二维矩阵表示一个图像。将图像顺时针旋转 90 度。说明：你必须在原地旋转图像，这意味着你需要直接修改输入的二维矩阵。请不要使用另一个矩阵来旋转图像。


## 解题思路

- 给出一个二维数组，要求顺时针旋转 90 度。
- 这一题比较简单，按照题意做就可以。这里给出 2 种旋转方法的实现，顺时针旋转和逆时针旋转。

```c
    /*
     * clockwise rotate 顺时针旋转
     * first reverse up to down, then swap the symmetry 
     * 1 2 3     7 8 9     7 4 1
     * 4 5 6  => 4 5 6  => 8 5 2
     * 7 8 9     1 2 3     9 6 3
    */
    void rotate(vector<vector<int> > &matrix) {
        reverse(matrix.begin(), matrix.end());
        for (int i = 0; i < matrix.size(); ++i) {
            for (int j = i + 1; j < matrix[i].size(); ++j)
                swap(matrix[i][j], matrix[j][i]);
        }
    }
    
    /*
     * anticlockwise rotate 逆时针旋转
     * first reverse left to right, then swap the symmetry
     * 1 2 3     3 2 1     3 6 9
     * 4 5 6  => 6 5 4  => 2 5 8
     * 7 8 9     9 8 7     1 4 7
    */
    void anti_rotate(vector<vector<int> > &matrix) {
        for (auto vi : matrix) reverse(vi.begin(), vi.end());
        for (int i = 0; i < matrix.size(); ++i) {
            for (int j = i + 1; j < matrix[i].size(); ++j)
                swap(matrix[i][j], matrix[j][i]);
        }
    }
```