48. Rotate Image

You are given an n x n 2D matrix representing an image.

Rotate the image by 90 degrees (clockwise).

顺时针旋转

/*
 * clockwise rotate
 * 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
*/

class Solution {
public:
    X.
};

逆时针旋转

/*
 * anticlockwise rotate
 * 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]);
    }
}

直观的方法，一圈一圈的旋转

class Solution {
public:
    void rotateEdge(vector<vector<int>> &matrix, int sR, int sC, int dR, int dC){
        int len = dR - sR;
        for(int i = 0; i < len; i++){
            int tmp = matrix[sR][sC + i];

            matrix[sR][sC + i] = matrix[dR - i][sC];
            matrix[dR - i][sC] = matrix[dR][dC - i];
            matrix[dR][dC - i] = matrix[sR + i][dC];
            matrix[sR + i][dC] = tmp;
        }
    }

    void rotate(vector<vector<int>> &m){
        int n = m.size();

        int sR = 0, sC = 0;
        int eR = n - 1, eC = n - 1;


        while(sR < eR){
            rotateEdge(m, sR++, sC++, eR--, eC--);
        }
    }
};