将图像旋转 90 度 · GeeksForGeeks 中文系列教程

# 将图像旋转 90 度 > 原文： [https://www.geeksforgeeks.org/turn-an-image-by-90-degree/](https://www.geeksforgeeks.org/turn-an-image-by-90-degree/) 给定图像，如何将其旋转 90 度？一个模糊的问题。最小化浏览器，然后尝试进一步解决方案。图像可以视为 2D 矩阵，可以存储在缓冲区中。我们提供了矩阵尺寸及其基地址。我们怎么能转呢？例如，见下图， ``` * * * ^ * * * * * * | * * * * * * | * * * * * * | * * * ``` 向右旋转后，它会出现（观察箭头方向） ``` * * * * * * * * * * * * -- - - > * * * * * * * * * * * * ``` 这个想法很简单。将源矩阵的每一行转换为最终图像所需的列。我们将使用辅助缓冲区来转换图像。从上图可以看出 ``` first row of source ------> last column of destination second row of source ------> last but-one column of destination so ... on last row of source ------> first column of destination ``` 以图片的形式，我们可以表示（m x n）矩阵到（n x m）矩阵的上述转换， ![](https://img.kancloud.cn/d0/ad/d0adb58af69d2e3cf7856aaf5d5c7c40_541x223.png) 转变如果您尚未尝试，请立即尝试使用伪代码。编写伪代码很容易。在 C/C++ 中，我们通常会按行主要顺序遍历矩阵。每行转换为最终图像的不同列。我们需要构造最终图像的列。请参阅以下算法（转换） ``` for (r = 0; r < m; r++) { for (c = 0; c < n; c++) { // Hint: Map each source element indices into // indices of destination matrix element. dest_buffer [ c ] [ m - r - 1 ] = source_buffer [ r ] [ c ]; } } ``` 请注意，有多种方法可以基于矩阵，行主要或列主要顺序的遍历来实现算法。我们有两个矩阵和两种方式（行和列主行）遍历每个矩阵。因此，至少有 4 种不同的方式将源矩阵转换为最终矩阵。 ## C++ ```cpp // C++ program to turn an // image by 90 Degree #include <bits/stdc++.h> using namespace std; void displayMatrix(unsigned int const *p, unsigned int row, unsigned int col); void rotate(unsigned int *pS, unsigned int *pD, unsigned int row, unsigned int col); void displayMatrix(unsigned int const *p, unsigned int r, unsigned int c) { unsigned int row, col; cout << "\n\n"; for (row = 0; row < r; row++) { for (col = 0; col < c; col++) cout << * (p + row * c + col) << "\t"; cout << "\n"; } cout << "\n\n"; } void rotate(unsigned int *pS, unsigned int *pD, unsigned int row, unsigned int col) { unsigned int r, c; for (r = 0; r < row; r++) { for (c = 0; c < col; c++) { *(pD + c * row + (row - r - 1)) = *(pS + r * col + c); } } } // Driver Code int main() { // declarations unsigned int image[][4] = {{1, 2, 3, 4}, {5, 6, 7, 8}, {9, 10, 11, 12}}; unsigned int *pSource; unsigned int *pDestination; unsigned int m, n; // setting initial values // and memory allocation m = 3, n = 4, pSource = (unsigned int *)image; pDestination = (unsigned int *)malloc (sizeof(int) * m * n); // process each buffer displayMatrix(pSource, m, n); rotate(pSource, pDestination, m, n); displayMatrix(pDestination, n, m); free(pDestination); return 0; } // This code is contributed by rathbhupendra ``` ## C ``` // C program to turn an // image by 90 Degree #include <stdio.h> #include <stdlib.h> void displayMatrix(unsigned int const *p, unsigned int row, unsigned int col); void rotate(unsigned int *pS, unsigned int *pD, unsigned int row, unsigned int col); void displayMatrix(unsigned int const *p, unsigned int r, unsigned int c) { unsigned int row, col; printf("\n\n"); for (row = 0; row < r; row++) { for (col = 0; col < c; col++) printf("%d\t", * (p + row * c + col)); printf("\n"); } printf("\n\n"); } void rotate(unsigned int *pS, unsigned int *pD, unsigned int row, unsigned int col) { unsigned int r, c; for (r = 0; r < row; r++) { for (c = 0; c < col; c++) { *(pD + c * row + (row - r - 1)) = *(pS + r * col + c); } } } // Driver Code int main() { // declarations unsigned int image[][4] = {{1,2,3,4}, {5,6,7,8}, {9,10,11,12}}; unsigned int *pSource; unsigned int *pDestination; unsigned int m, n; // setting initial values // and memory allocation m = 3, n = 4, pSource = (unsigned int *)image; pDestination = (unsigned int *)malloc (sizeof(int) * m * n); // process each buffer displayMatrix(pSource, m, n); rotate(pSource, pDestination, m, n); displayMatrix(pDestination, n, m); free(pDestination); getchar(); return 0; } ``` 输出： ``` 1 2 3 4 5 6 7 8 9 10 11 12 9 5 1 10 6 2 11 7 3 12 8 4 ``` [将方形矩阵旋转 90 度](https://www.geeksforgeeks.org/inplace-rotate-square-matrix-by-90-degrees/) 由 [**Venki**](https://www.geeksforgeeks.org/?page_id=2) 编译。如果发现任何不正确的地方，或者想分享有关上述主题的更多信息，请写评论。