每日算法之三十七：Rotate Image （图像旋转）

来源：互联网发布：绝地求生知乎编辑：程序博客网时间：2024/05/16 10:17

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。因为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++)//每次能够交换四个元素的位置         {             int temp = matrix[i][layer];//不清楚画图举例即可             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;          }    }};

下面是网友给出的另一种方案：

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;            }        }    }};

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