# 【LeetCode】 48. Rotate Image

## 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, which means you have to modify the input 2D matrix directly. DO NOT allocate another 2D matrix and do the rotation.

> 給你一個n*n的二維矩陣，代表一張圖片。
> 請順時鐘旋轉該矩陣九十度。
>
> 注意：
> 你必須使用in-place演算法去做旋轉，也就是說你必須直接修改輸入的二維矩陣。不要去分配其他的二維矩陣去做旋轉。

## Example:

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]
]

## Solution

* 先將矩陣轉置，再將每一個column反轉即可。
* 因為規定使用in-place，所以基本上function都要用交換去完成。
* 將vector.size存起來可以稍微加速。

### Code

class Solution {
public:
    
    void Exchange(int& a,int& b)
    {
        int temp = a;
        a = b;
        b = temp;
    }
    
    void Transpose(vector<vector<int>>& matrix)
    {
        int dimention = matrix.size();
        for(int i = 0;i<dimention;i++)
        {
            for(int j = i+1;j<dimention;j++)
                Exchange(matrix[i][j],matrix[j][i]);
        }
    }
    
    void Reserve(vector<int>& v)
    {
        int s = v.size();
        for(int i = 0;i<s/2;i++)
            Exchange(v[i],v[s-1-i]);
    }
    
    void rotate(vector<vector<int>>& matrix) {
        Transpose(matrix);
        for(int i = 0;i<matrix.size();i++)
            Reserve(matrix[i]);
    }
};

###### tags: `LeetCode` `C++`