每日算法37：Rotate Image （图像旋转）

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

Rotate the image by 90 degrees (clockwise).

Follow up:

Could you do this in-place?

原地图像顺时针旋转90度。由于要求空间复杂度是常数，因此应该迭代旋转操作。

class Solution {
public:
void rotate(vector<vector<int> > &matrix) {
int n = matrix.size();
int layers = n/2;//图像旋转的圈数

for(int layer = 0;layer < layers;layer++)//每次循环一层，右青色到紫色
for(int i = layer;i<n-1-layer;i++)//每次可以交换四个元素的位置
{
int temp = matrix[i][layer];//不清楚绘图举例就可以
matrix[i][layer] = matrix[n-1-layer][i];
matrix[n-1-layer][i] = matrix[n-1-i][n-1-layer];
matrix[n-1-i][n-1-layer] = matrix[layer][n-1-i];
matrix[layer][n-1-i] = temp;
}
}
};

以下是网友给出的还有一种方案：

class Solution {
public:
void rotate(vector<vector<int> > &matrix) {
int i,j,temp;
int n=matrix.size();
// 沿着副对角线反转
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n - i; ++j) {
temp = matrix[i][j];
matrix[i][j] = matrix[n - 1 - j][n - 1 - i];
matrix[n - 1 - j][n - 1 - i] = temp;
}
}
// 沿着水平中线反转
for (int i = 0; i < n / 2; ++i){
for (int j = 0; j < n; ++j) {
temp = matrix[i][j];
matrix[i][j] = matrix[n - 1 - i][j];
matrix[n - 1 - i][j] = temp;
}
}
}
};