假设我有一个以0°旋转存储的数组：

0 0 1 0 0
0 0 1 0 0 
1 1 1 0 0 
0 0 0 0 0
0 0 0 0 0 

如果我通过,我希望它以良好的近似值返回,例如30°作为参数,它将是这样的：

0 0 0 1 0
1 1 0 1 0
0 0 1 0 0 
0 0 0 0 0 
0 0 0 0 0 

45°会

1 0 0 0 1
0 1 0 1 0
0 0 1 0 0 
0 0 0 0 0 
0 0 0 0 0

I am aware of the solutions posted for 90° rotations.但我认为这不会对我有帮助吗？

我没有任何示例代码,因为我目前甚至不知道从哪里开始查找.如果有任何关键字我可以google指向我的方向我可以适应这个,这也将是伟大的.

Spectre在C#中的代码解决方案：

    class Rotation
{

    public Rotation() {
        A = new int[xs,ys]{
{0,9,0},{0,{9,};
        B = new int[xs,ys];

        deg = (float)(Math.PI / 180.0);
    }

    public const int xs = 7; // matrix size
    public const int ys = 7;
    const int x0 = 3; // rotation center cell
    const int y0 = 3;
    readonly float deg; 
    public int[,] A;
    public int[,] B;

    //---------------------------------------------------------------------------

    public void rotcv(float ang) {
        rotcw(Rotation.x0,Rotation.y0,ang);
    }
    private void rotcw(int x0,int y0,float ang) // rotate A -> B by angle ang around (x0,y0) CW if ang>0
    {
        int x,y,ix0,iy0,ix1,iy1,q;
        double xx,yy,fx,fy,c,s;
        // circle kernel
        c = Math.Cos(-ang); s = Math.Sin(-ang);
        // rotate
        for (y = 0; y < ys; y++)
            for (x = 0; x < xs; x++)
            {
                // offset so (0,0) is center of rotation
                xx = x - x0;
                yy = y - y0;
                // rotate (fx,fy) by ang
                fx = ((xx * c) - (yy * s));
                fy = ((xx * s) + (yy * c));
                // offset back and convert to ints and weights
                fx += x0; ix0 = (int)Math.Floor(fx); fx -= ix0; ix1 = ix0 + 1; if (ix1 >= xs) ix1 = ix0;
                fy += y0; iy0 = (int)Math.Floor (fy); fy -= iy0; iy1 = iy0 + 1; if (iy1 >= ys) iy1 = iy0;
                // bilinear interpolation A[fx][fy] -> B[x][y]
                if ((ix0 >= 0) && (ix0 < xs) && (iy0 >= 0) && (iy0 < ys))
                {
                    xx = (A[ix0,iy0]) + ((A[ix1,iy0] - A[ix0,iy0]) * fx);
                    yy = (A[ix0,iy0]) * fx);
                    xx = xx + ((yy - xx) * fy); q =(int) xx;
                }
                else q = 0;
                B[x,y] = q;
            }
    }
}

测试：

 static void Main(string[] args)
    {
        Rotation rot = new Rotation();

        for (int x = 0; x < Rotation.xs; x++) {
            for (int y = 0; y < Rotation.xs; y++) {
                Console.Write(rot.A[x,y] + " ");
            }
            Console.WriteLine();
        }
        Console.WriteLine();
        float rotAngle = 0;
        while (true)
        {
            rotAngle += (float)(Math.PI/180f)*90;
            rot.rotcv(rotAngle);
            for (int x = 0; x < Rotation.xs; x++)
            {
                for (int y = 0; y < Rotation.xs; y++)
                {
                    Console.Write(rot.B[x,y] + " ");
                }
                Console.WriteLine();
            }
            Console.WriteLine();
            Console.ReadLine();
        }

    }