Question

我正在尝试定义一个简单的Scala方法来确定两个矩形是否在任何容量中重叠。我相信这样做的方式将是：

if (! (Range_of_rectangle1.contains(anything in Range_of_rectangle2) ) ) => false

这需要针对x轴和y轴进行... 但我对Scala足够新，我不确定如何编写类似someRange.contains( anything in another Range)的内容。

我目前用于确定重叠分类的代码有效但有一些问题，我将讨论：

首先我定义一个Rectangle（并且我的代码中有一个case class的原因，但这与此任务无关）。

case class Rectangle (minx: Int, maxx: Int, miny: Int, maxy: Int)

然后我创建函数来查看两个矩形是否重叠

def rectanglesOverlap(r1: Rectangle, r2:Rectangle): Boolean = {
  r2 match {
     //In English: if r2's minx OR miny are not anywhere in the range of r1's x-axis, then there's no overlap along the x-axis
     //If the range of r1's x-axis does NOT contain anything from r2's x-axis, they don't overlap
     case x_overlap1 if (! (  (r1.minx to r1.maxx).contains(r2.minx) || (r1.minx to r1.maxx).contains(r2.maxx) ) ) => false //where r1 is larger rectangle 
     case y_overlap1 if (! (  (r1.miny to r1.maxy).contains(r2.miny) || (r1.miny to r1.maxy).contains(r2.maxy) ) ) => false
     //If the range of r2's x-axis does NOT contain anything from r1's x-axis, they don't overlap
     case x_overlap2 if (! (  (r2.minx to r2.maxx).contains(r1.minx) || (r2.minx to r2.maxx).contains(r1.maxx) ) ) => false //where r2 is larger rectangle
     case y_overlap2 if (! (  (r2.miny to r2.maxy).contains(r1.miny) || (r2.miny to r2.maxy).contains(r1.maxy) ) ) => false
     case _ => true
  }
}

那么代码尝试做的是从一个矩形的x轴和y轴开始，并检查另一个矩形的minx / y OR maxx / y是否在那里....

看到问题？

当我测试它时，我得到了“假”：

val q1 = Rectangle(1, 18, 1, 18)  
val q2 = Rectangle(1,8,8,16) 
scala> rectanglesOverlap(q1, q2) 
res0: Boolean = false

原因很明显。它是错误的，因为q2 y轴是8-16，并且q1 miny（1）或q1 maxy（18）都不在8-16范围内。但是，很明显它们重叠。

所以要知道我在概念上知道我的代码有什么问题，我试图弄清楚如何以编程方式执行这样的操作：

someRange.contains( anything in another Range)。

但是我搜索Google和Stack Overflow的努力还没有得到正确的解决方案。帮助

Answer 1

当你想知道一个集合与另一个集合重叠的地方时，你正在寻找他们的“交集”。

someRange.intersect(anotherRange)

或

(1 to 18) intersect (8 to 16)

并将其变为布尔

((1 to 18) intersect (8 to 16)).nonEmpty

Scala Range包含（其他范围内的任何内容）？

1 个答案: