将图像原地旋转 90 度的算法?(没有额外的内存) [英] Algorithm to rotate an image 90 degrees in place? (No extra memory)

问题描述

在嵌入式 C 应用程序中，我有一个大图像，我想将其旋转 90 度.目前我使用众所周知的简单算法来做到这一点.但是，该算法要求我制作另一个图像副本.我想避免为副本分配内存，我宁愿就地旋转它.由于图像不是方形的，这很棘手.有人知道合适的算法吗?

In an embedded C app, I have a large image that I'd like to rotate by 90 degrees. Currently I use the well-known simple algorithm to do this. However, this algorithm requires me to make another copy of the image. I'd like to avoid allocating memory for a copy, I'd rather rotate it in-place. Since the image isn't square, this is tricky. Does anyone know of a suitable algorithm?

编辑添加澄清，因为人们在问:

Edited to add clarification, because people are asking:

我以通常的格式存储图像:

I store an image in the usual format:

// Images are 16 bpp
struct Image {
    int width;
    int height;
    uint16_t * data;
};

uint16_t getPixel(Image *img, int x, int y)
{
    return img->data[y * img->width + x];
}

我希望移动 data 数组的内容，然后交换 width 和 height 成员变量.所以如果我从一个 9x20 像素的图像开始，然后旋转它，我最终会得到一个 20x9 像素的图像.这改变了图像的步幅，使算法复杂化了很多.

I'm hoping to move the contents of the data array around, then swap over the width and height member variables. So if I start with a 9x20 pixel image, then rotate it, I'll end up with a 20x9 pixel image. This changes the stride of the image, which complicates the algorithm a lot.

推荐答案

这可能会有所帮助:In-位矩阵转置.

(如 rlbond 所述，您可能还需要在移调后进行一些镜像).

(You might also have to do some mirroring after the transposition, as rlbond mentions).

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