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, 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]
]

分析

旋转的时候，元素的交换只发生在所有相距45度的4个元素之间。而这些包含4个元素的小组，可以按照矩阵的层次由内而外分成[0, (n+1)./2)层，而每一层中包含4[0, n-2i-1)个元素。这样遍历每一组，然后交换每一组中的元素即可。

实现

class Solution {
public:
    void rotate(vector<vector<int>>& matrix) {
        int n=matrix.size();
        for(int i=0; i<(n+1)/2; i++){
            for(int j=0; j<n-2*i-1; j++){
                int x1=i, y1=i+j;
                int x2=i+j, y2=n-1-i;
                int x3=n-1-i, y3=n-1-(i+j);
                int x4=n-1-(i+j), y4=i;
                int tmp = matrix[x4][y4];
                matrix[x4][y4] = matrix[x3][y3];
                matrix[x3][y3] = matrix[x2][y2];
                matrix[x2][y2] = matrix[x1][y1];
                matrix[x1][y1] = tmp;
            }
        }
    }
};

思考

我这种方法应该是比较直接的。另外取巧的方法还有，将转置分成两步完成：首先沿副对角线翻转，再沿水平中心线翻转即可。或者先沿水平中线翻转，再沿主对角线翻转。