Android中的點內多邊形測試

Question

有一天我在Java中做了一個類來計算point(X,Y)是否在多邊形內。 （X和Y是double，因爲將是地理座標）。

我知道Java有類Polygon，但我不得不使用Path2D和Point2D，因爲Polygon不允許double的，只是整數:(

我一旦多邊形在Path2D做的，我使用的方法contains（Path2D了吧），我的問題解決了

但現在，我要導入到Android的，問題就在這裏，因爲Path2D需要進口。

import java.awt.geom.Path2D; 
import java.awt.geom.Point2D; 

在Android中不存在awt，所以我無法使用。

那麼，有沒有類似於Path2D那有contains方法？或者我必須自己計算？

這是我如何使用Path2D沒有在Java中：

private void ConstructPolygon(Vector<Point2D> coodinates) 
{  
    this.polygon.moveTo(coodinates.get(0).getX(), coodinates.get(0).getY());   

    //System.out.println(coodinates.get(0).getX() + " " + coodinates.get(0).getY()); 
    //System.out.println("asda"); 

    for(int i = 1; i < this.num_points; i++) 
    { 
     //System.out.println(coodinates.get(i).getX() + " " + coodinates.get(i).getY()); 
     this.polygon.lineTo(coodinates.get(i).getX(), coodinates.get(i).getY()); 
    } 
    this.polygon.closePath(); 
} 
public boolean InsideCity(Point2D punto) 
{ 
    return this.polygon.contains(punto);     
} 

Answer 1

您可以準確地用我的簡單的庫這樣的：https://github.com/snatik/polygon-contains-point。

準備多邊形：

Polygon polygon = Polygon.Builder() 
    .addVertex(new Point(1, 3)) 
    .addVertex(new Point(2, 8)) 
    .addVertex(new Point(5, 4)) 
    .addVertex(new Point(5, 9)) 
    .addVertex(new Point(7, 5)) 
    .addVertex(new Point(6, 1)) 
    .addVertex(new Point(3, 1)) 
    .build(); 

並檢查，而該點位於多邊形內：

Point point = new Point(4.5f, 7); 
boolean contains = polygon.contains(point); 

它與浮動的類型，幷包含孔:)

Answer 2

對不起多邊形@ sromku我問自己（我從來沒有使用這種類型的東西）

這就是我如何解決如果任何人有相同的問題

Builder poly2 = new Polygon.Builder(); 
    for(int i = 0; i< xpoints.length;i++){ 
     poly2.addVertex(new Point(xpoints[i],ypoints[i])); 
    } 
    Polygon polygon2 = poly2.build(); 

Answer 3

這是我在Android中做到的。 正是基於這個java程序（光線投射算法）： https://gis.stackexchange.com/questions/42879/check-if-lat-long-point-is-within-a-set-of-polygons-using-google-maps/46720#46720

public boolean pointInPolygon(LatLng point, Polygon polygon) { 
     // ray casting alogrithm http://rosettacode.org/wiki/Ray-casting_algorithm 
     int crossings = 0; 
     List<LatLng> path = polygon.getPoints(); 
     path.remove(path.size()-1); //remove the last point that is added automatically by getPoints() 

     // for each edge 
     for (int i=0; i < path.size(); i++) { 
      LatLng a = path.get(i); 
      int j = i + 1; 
      //to close the last edge, you have to take the first point of your polygon 
      if (j >= path.size()) { 
       j = 0; 
      } 
      LatLng b = path.get(j); 
      if (rayCrossesSegment(point, a, b)) { 
       crossings++; 
      } 
     } 

     // odd number of crossings? 
     return (crossings % 2 == 1); 
    } 

    public boolean rayCrossesSegment(LatLng point, LatLng a,LatLng b) { 
       // Ray Casting algorithm checks, for each segment, if the point is 1) to the left of the segment and 2) not above nor below the segment. If these two conditions are met, it returns true 
       double px = point.longitude, 
       py = point.latitude, 
       ax = a.longitude, 
       ay = a.latitude, 
       bx = b.longitude, 
       by = b.latitude; 
      if (ay > by) { 
       ax = b.longitude; 
       ay = b.latitude; 
       bx = a.longitude; 
       by = a.latitude; 
      } 
      // alter longitude to cater for 180 degree crossings 
      if (px < 0 || ax <0 || bx <0) { px += 360; ax+=360; bx+=360; } 
      // if the point has the same latitude as a or b, increase slightly py 
      if (py == ay || py == by) py += 0.00000001; 


      // if the point is above, below or to the right of the segment, it returns false 
      if ((py > by || py < ay) || (px > Math.max(ax, bx))){ 
       return false; 
      } 
      // if the point is not above, below or to the right and is to the left, return true 
      else if (px < Math.min(ax, bx)){ 
       return true; 
      } 
      // if the two above conditions are not met, you have to compare the slope of segment [a,b] (the red one here) and segment [a,p] (the blue one here) to see if your point is to the left of segment [a,b] or not 
      else { 
       double red = (ax != bx) ? ((by - ay)/(bx - ax)) : Double.POSITIVE_INFINITY; 
       double blue = (ax != px) ? ((py - ay)/(px - ax)) : Double.POSITIVE_INFINITY; 
       return (blue >= red); 
      } 

    } 

Answer 4

您可以使用谷歌地圖PolyUtil：

import com.google.maps.android.PolyUtil; 

boolean inside = PolyUtil.containsLocation(new LatLng(...), poly, true);