算法精粹(algorithm-essentials)

感谢soulmachine@github提供内容
## Rotate Image


### 描述

You are given an `n × n` 2D matrix representing an image.

Rotate the image by 90 degrees (clockwise).

Follow up:
Could you do this in-place?


### 分析

首先想到,纯模拟,从外到内一圈一圈的转,但这个方法太慢。

如下图,首先沿着副对角线翻转一次,然后沿着水平中线翻转一次。

![Rotate image](../../images/rotate-image.png)

或者,首先沿着水平中线翻转一次,然后沿着主对角线翻转一次。


### 代码1

{% if book.java %}
{% codesnippet "./code/rotate-image-1."+book.suffix, language=book.suffix %}{% endcodesnippet %}
{% endif %}

{% if book.cpp %}
```cpp
// Rotate Image
// 思路 1,时间复杂度O(n^2),空间复杂度O(1)
class Solution {
public:
void rotate(vector>& matrix) {
const int n = matrix.size();

for (int i = 0; i < n; ++i) // 沿着副对角线反转
for (int j = 0; j < n - i; ++j)
swap(matrix[i][j], matrix[n - 1 - j][n - 1 - i]);

for (int i = 0; i < n / 2; ++i) // 沿着水平中线反转
for (int j = 0; j < n; ++j)
swap(matrix[i][j], matrix[n - 1 - i][j]);
}
};
```
{% endif %}


### 代码2

{% if book.java %}
{% codesnippet "./code/rotate-image-2."+book.suffix, language=book.suffix %}{% endcodesnippet %}
{% endif %}

{% if book.cpp %}
```cpp
// Rotate Image
// 思路 2,时间复杂度O(n^2),空间复杂度O(1)
class Solution {
public:
void rotate(vector>& matrix) {
const int n = matrix.size();

for (int i = 0; i < n / 2; ++i) // 沿着水平中线反转
for (int j = 0; j < n; ++j)
swap(matrix[i][j], matrix[n - 1 - i][j]);

for (int i = 0; i < n; ++i) // 沿着主对角线反转
for (int j = i + 1; j < n; ++j)
swap(matrix[i][j], matrix[j][i]);
}
};
```
{% endif %}