leetcode 48. Rotate Image

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本文介绍了一种不使用额外空间直接修改输入二维矩阵以实现顺时针旋转90度的方法。通过两步操作：首先沿主对角线交换元素，然后沿中轴线交换每列元素，实现矩阵的高效旋转。

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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.

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]
]

方法一：纯找数学规律，按照相应应该在的位置安放

方法二：
本地使得二维矩阵，旋转90角度。
通过实际数据分析，通过两个步骤的元素交换可实现目标：
1\按照主对角线，将对称元素交换
2\按照列，将对称列元素全部交换
即可达到，使得二维矩阵，本地旋转90个角度。

class Solution {
public:
    void jiaohuan(vector<vector<int>>& matrix, int i, int h)
    {
         int n = matrix.size();//4
         int temp;
         temp = matrix[i][h];//0 1
         matrix[i][h] = matrix[n-1-h][i];//20 ->01
         matrix[n-1-h][i] = matrix[n-1-i][n-1-h];// 32 ->20
         matrix[n-1-i][n-1-h] = matrix[h][n-1-i];
         matrix[h][n-1-i] = temp;
    }
    
    void rotate(vector<vector<int>>& matrix) 
    {
        //way-1
        /*
        int n = matrix.size();
        int h, cishu;
        for (int i = 0; i < n/2; i++)
        {
            h = i;
            cishu = (n - 1) - 2 * i;
            while (cishu > 0)
            {
                jiaohuan(matrix, i, h);
                h++;
                cishu--;
            }
        }
        */
        
        //way-2
        
        int n = matrix.size();
        for (int i = 0; i < n; i++)
            for (int j = 0; j < i; j++)
                swap(matrix[i][j], matrix[j][i]);
        
        for (int i = 0; i < n; i++)
            for (int j = 0; j < n/2; j++)
                swap(matrix[i][j], matrix[i][n-1-j]);  
    }
};