java - 二维坐标上的流/IntRange？

Question

我有以下类(class):

public class Rectangle
{
   public int width, height;
   public Point start;

   public Rectangle(Point start, int width, int height)
   {
      this.start = start;
      this.width = width;
      this.height = height;
   }
}

public class Point
{
   public int x, y;

   public Point(int x, int y)
   {
      this.x = x;
      this.y = y;
   }
}

我想创建一个 Rectangle 类的方法，以便返回类型为 Point 的 Java 8 Stream，它将流式传输矩形内的索引位置。前任。

// This is inside the rectangle class.
public Stream<Point> stream(int hstep, int vstep)
{
   // if the rectangle started at 0,0
   // and had a width and height of 100
   // and hstep and vstep were both 1
   // then the returned stream would be
   // 0,0 -> 1,0 -> 2,0 -> ... -> 100,0 -> 0,1 -> 1,1 -> ...

   // If vstep was 5 and h step was 25 it would return
   // 0,0 -> 25,0 -> 50,0 -> 75,0 -> 100,0 -> 0,5 -> 25,5 -> ...

   return ...
}

我经常使用 IntStream、map、filter 等，但这比我尝试过的任何东西都复杂得多.我不知道我会怎么做那样的事情。有人可以引导我走向正确的方向吗？

Answer 1

您可以使用嵌套的 IntStream 生成每个 Point，然后展平生成的流:

public Stream<Point> stream(int hstep, int vstep) {
    return IntStream.range(0, height / vstep)
            .mapToObj(y -> IntStream.range(0, width / hstep)
                    .mapToObj(x -> new Point(start.x + x * hstep, start.y + y * vstep)))
            .flatMap(Function.identity());
}