坐标算法 - 围绕中心旋转

Coordinate Algorithm - Rotate around the center(坐标算法 - 围绕中心旋转)

问题描述

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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