C++ 将 2D 像素阵列旋转 90 度

Rotating a 2D pixel array by 90 degrees

提问by noob

I have an array of pixel data for an image. The image I am getting is already rotated to 270 degrees. So I am trying to rotate it again by 90 degrees to have the correct image. I've tried a transpose algorithm, by changing data[x][y]to data[y][x], but I don't think that's the correct way. Can anyone guide me what can I do to have it rotated?

我有一个图像的像素数据数组。我得到的图像已经旋转到 270 度。所以我试图将它再次旋转 90 度以获得正确的图像。我已经尝试了转置算法，通过更改data[x][y]为data[y][x]，但我认为这不是正确的方法。谁能指导我我该怎么做才能让它旋转？

回答by raj raj

You have old_data[rows][cols]and new_data[cols][rows], then:

你有old_data[rows][cols]和new_data[cols][rows]，那么：

for(int i=0; i<cols; i++) {
    for(int j=0; j<rows; j++) {
        new_data[i][j] = old_data[rows-1-j][i];
    }
}

This should rotate old_data by 90 degrees CW.

这应该将 old_data 旋转 90 度 CW。

回答by mhaghighat

If you want to do it in-place with O(1) space, you can follow this:

如果你想用 O(1) 空间就地做，你可以按照这个：

1. Transpose the matrix by swapping data[i][j]and data[j][i]:

   for (int i = 0; i < n; i += 1) {
       for (int j = i+1; j < n; j += 1) {
           swap(data[i][j], data[j][i]);
       }
   }
   

2. Reverse each row or columnfor +90 or -90 degrees of rotation, respectively. For example for +90 degrees of rotation:

   for (int i = 0; i < n; i += 1) {
       for (int j = 0; j < n/2; j += 1) {
           swap(data[i][j], data[i][n-1-j]);
       }
   }
   

1. 通过交换data[i][j]和 来转置矩阵data[j][i]：

   for (int i = 0; i < n; i += 1) {
       for (int j = i+1; j < n; j += 1) {
           swap(data[i][j], data[j][i]);
       }
   }
   

2. 分别反转每行或每列+90 或 -90 度的旋转。例如对于 +90 度旋转：

   for (int i = 0; i < n; i += 1) {
       for (int j = 0; j < n/2; j += 1) {
           swap(data[i][j], data[i][n-1-j]);
       }
   }
   

回答by herohuyongtao

This can be done without using any extra space, so called In-place matrix transposition(not exact the same). Remember to do some mirroring after the transposition.

这可以在不使用任何额外空间的情况下完成，所谓的就地矩阵转置（不完全相同）。记得在换位后做一些镜像。

1. If the image is square

   

2. If the image is not square

   • For non-square matrices, the algorithms are more complicated. Many of the algorithms prior to 1980 could be described as "follow-the-cycles" algorithms. That is, they loop over the cycles, moving the data from one location to the next in the cycle. In pseudocode form:

   

1. 如果图像是正方形

   

2. 如果图像不是方形的

   • 对于非方阵，算法更复杂。1980 年之前的许多算法都可以被描述为“follow-the-cycles”算法。也就是说，它们在循环中循环，将数据从循环中的一个位置移动到下一个位置。伪代码形式：

   

回答by Akinjiola Toni

To rotate the image (2D matrix) by 90deg, you can easily do this by mapping out a pattern between the initial state and the end state after rotating it by 90deg.

要将图像（2D 矩阵）旋转 90 度，您可以通过在旋转 90 度后绘制初始状态和结束状态之间的模式来轻松完成此操作。

---

---

a[i][j] => a[m][n]
a[0][0] => a[0][2]
a[0][1] => a[1][2]
a[0][2] => a[2][2]
a[1][0] => a[0][1]
a[1][1] => a[1][1]
a[1][2] => a[2][1]
a[2][0] => a[0][0]
a[2][1] => a[1][0]
a[2][2] => a[2][0]

Now the solution is obvious. All the J's turn to M and N = (Size of matrix(2) - I).

现在解决方案是显而易见的。所有的 J 都转向 M 和 N = (矩阵的大小 (2) - I)。

---

---

const rotateImage = (a) => {
  let size = a.length;
  let results = new Array(size);
  for (let i = 0; i < size; i++) {
    results[i] = new Array(size);
  }
  for (let i = 0; i < size; i++) {
    for (let j = 0; j < size; j++) {
      results[j][(size - 1) - i] = a[i][j];
    }
  }
  return results;
}

console.log(rotateImage([
  [1, 2, 3],
  [4, 5, 6],
  [7, 8, 9]
]));