坐标算法 - 围绕中心旋转
通过查看这张图片,我想你会很好地理解我的问题:
By looking at this image I think you will understand my problem pretty well:
(图片已删除 - 网址不再有效,现在返回广告)
所以基本上我想要一个以对象为参数的函数,并根据我之前添加的对象数量为该对象提供正确的坐标.
So basically I want a function that takes an object as parameter and gives this object the correct coordinates based on how many objects I've added before.
假设我会将所有这些对象添加到一个数组中:
Let's say I would add all these objects to an array:
objectArray[]
每次我添加一个新对象时:objectArray.add(object)
Each time I add a new object:
objectArray.add(object)
object.x
和 object.y
坐标将基于某种算法设置:
The object.x
and object.y
coordinates will be set based on some algorithm:
object.x = ?
object.y = ?
(我正在使用 Java)
(I'm working in Java)
感谢您的帮助.
推荐答案
这是不依赖循环的封闭式解决方案...我对 Java 不太熟悉,所以它在 C# 中,但它使用基本操作.
Here's the closed-form solution that doesn't rely on a loop... I'm not handy with Java, so it's in C#, but it uses basic operations.
static void SpiralCalc(int i) {
i -= 2;
// Origin coordinates
int x = 100, y = 100;
if (i >= 0) {
int v = Convert.ToInt32(Math.Truncate(Math.Sqrt(i + .25) - .5));
int spiralBaseIndex = v * (v + 1);
int flipFlop = ((v & 1) << 1) - 1;
int offset = flipFlop * ((v + 1) >> 1);
x += offset; y += offset;
int cornerIndex = spiralBaseIndex + (v + 1);
if (i < cornerIndex) {
x -= flipFlop * (i - spiralBaseIndex + 1);
} else {
x -= flipFlop * (v + 1);
y -= flipFlop * (i - cornerIndex + 1);
}
}
// x and y are now populated with coordinates
Console.WriteLine(i + 2 + " " + x + " " + y);
}
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