java - 如何编写带有 Axis 的递归算法？

Question

我需要递归地在 x 和 y Axis 坐标上找到一个对象。在这段特定的代码中，我需要通过重复调用搜索方法来找到位于坐标 168,250 中的“Remy”

import java.awt.Point;

public class Sensor {

private Point target;
private Point[] points = {new Point(378, 349), new Point(147, 315), new Point(95, 375), new Point(242, 493), new Point(379, 389), new Point(168, 250), new Point(130, 220), new Point(160, 200), new Point(0, 0), new Point(0, 511), new Point(511, 0), new Point(511, 511)};
private String[] names = {"Bernstein", "DukeDog", "Huey", "Hedwig", "Flipper", "Remy", "QuadCat", "Nemo", "UL", "LL", "UR", "LR"};

public Sensor(String paramString) {
    this.target = new Point(512, 512);
    for (int i = 0; i < this.names.length; i++) {
        if (paramString.equalsIgnoreCase(this.names[i])) {
            this.target = this.points[i];
            break;
        }
    }
}

public static void main(String[] args) {
    Sensor sensor = new Sensor("Remy");
    Point result = sensor.search(0, 0, 512);
    if (result != null) {
        System.out.println("location: " + result.x + "," + result.y);
    } else {
        System.out.println("unable to find");
    }
}

public Point search(int x, int y, int width) {
    //Write a recursive algorithm to find Remy calling the scan method repeatedly such as: scan(x,y,width); 
    //todo 
    return null;
}

public int scan(int paramInt1, int paramInt2, int paramInt3) {
    if ((this.target.x >= paramInt1) && (this.target.x < paramInt1 + paramInt3) && (this.target.y >= paramInt2) && (this.target.y < paramInt2 + paramInt3)) {
        return paramInt3;
    } else {
        return -paramInt3;
    }

}

}

Answer 1

我建议这样做:

public Point search(int x, int y, int width) {

    System.out.println(x + "\t" + y + "\t" + width);

    int w = width/2;
    if(w<=1)
        return new Point(x,y); 
    int x1 = x+w;
    int y1 = y+w; 

    if(target.x>=x&&target.x<x1&&target.y>=y&&target.y<y1)
        return search(x,y,w); 
    else if(target.x>=x1&&target.x<x+w&&target.y>=y&&target.y<y1)
        return search(x1,y,w); 
    else if(target.x>=x&&target.x<x1&&target.y>=y1&&target.y<y1+w)
        return search(x,y1,w); 
    else /*if(target.x>=x1&&target.x<=x+w&&target.y>=y1&&target.y<=y1+w)*/
        return search(x1,y1,w); 
}

运行此方法时，输出为:

0   0   512
0   0   256
128 128 128
128 192 64
160 224 32
160 240 16
168 248 8
168 248 4
168 250 2
location: 168,250

请注意，它假设 width 是 2 的幂。这可以很容易地进行调整。

希望对您有所帮助。