尋找Path2D自相交

Question

我需要找到Path2D是否相交。現在，我只需從路徑中提取一系列行，然後查找是否存在任何這些行相交。但它具有O（n^2）的複雜性，所以它非常慢。有沒有更快的方法來做到這一點？

Answer 1

爲此，您可以更快地使用掃線算法：http://en.wikipedia.org/wiki/Sweep_line_algorithm

僞代碼：

Each line has a start point and an end point. Say that `start_x` <= `end_x` for all the lines. 
Create an empty bucket of lines. 
Sort all the points by their x coordinates, and then iterate through the sorted list. 
If the current point is a start point, test its line against all the lines in the bucket, and then add its line to the 
bucket. 
if the current point is an end point, remove its line from the bucket. 

最壞的情況是仍然O(N^2)，但平均情況O(NlogN)

Answer 2

這裏是我的該算法的Java實現：

import java.awt.Point; 
import java.awt.geom.Line2D; 
import java.awt.geom.PathIterator; 
import java.util.*; 

/** 
* Path2D helper functions. 
* <p/> 
* @author Gili Tzabari 
*/ 
public class Path2Ds 
{ 
    /** 
    * Indicates if a Path2D intersects itself. 
    * <p/> 
    * @return true if a Path2D intersects itself 
    */ 
    public static boolean isSelfIntersecting(PathIterator path) 
    { 
     SortedSet<Line2D> lines = getLines(path); 
     if (lines.size() <= 1) 
      return false; 

     Set<Line2D> candidates = new HashSet<Line2D>(); 
     for (Line2D line: lines) 
     { 
      if (Double.compare(line.getP1().distance(line.getP2()), 0) <= 0) 
      { 
       // Lines of length 0 do not cause self-intersection 
       continue; 
      } 
      for (Iterator<Line2D> i = candidates.iterator(); i.hasNext();) 
      { 
       Line2D candidate = i.next(); 

       // Logic borrowed from Line2D.intersectsLine() 
       int lineRelativeToCandidate1 = Line2D.relativeCCW(line.getX1(), line.getY1(), line. 
        getX2(), 
        line.getY2(), candidate.getX1(), candidate.getY1()); 
       int lineRelativeToCandidate2 = Line2D.relativeCCW(line.getX1(), line.getY1(), line. 
        getX2(), 
        line.getY2(), candidate.getX2(), candidate.getY2()); 
       int candidateRelativeToLine1 = Line2D.relativeCCW(candidate.getX1(), 
        candidate.getY1(), 
        candidate.getX2(), candidate.getY2(), line.getX1(), line.getY1()); 
       int candidateRelativeToLine2 = Line2D.relativeCCW(candidate.getX1(), 
        candidate.getY1(), 
        candidate.getX2(), candidate.getY2(), line.getX2(), line.getY2()); 
       boolean intersection = (lineRelativeToCandidate1 * lineRelativeToCandidate2 <= 0) 
        && (candidateRelativeToLine1 * candidateRelativeToLine2 <= 0); 
       if (intersection) 
       { 
        // Lines may share a point, so long as they extend in different directions 
        if (lineRelativeToCandidate1 == 0 && lineRelativeToCandidate2 != 0) 
        { 
         // candidate.P1 shares a point with line 
         if (candidateRelativeToLine1 == 0 && candidateRelativeToLine2 != 0) 
         { 
          // line.P1 == candidate.P1 
          continue; 
         } 
         if (candidateRelativeToLine1 != 0 && candidateRelativeToLine2 == 0) 
         { 
          // line.P2 == candidate.P1 
          continue; 
         } 
         // else candidate.P1 intersects line 
        } 
        else if (lineRelativeToCandidate1 != 0 && lineRelativeToCandidate2 == 0) 
        { 
         // candidate.P2 shares a point with line 
         if (candidateRelativeToLine1 == 0 && candidateRelativeToLine2 != 0) 
         { 
          // line.P1 == candidate.P2 
          continue; 
         } 
         if (candidateRelativeToLine1 != 0 && candidateRelativeToLine2 == 0) 
         { 
          // line.P2 == candidate.P2 
          continue; 
         } 
         // else candidate.P2 intersects line 
        } 
        else 
        { 
         // line and candidate overlap 
        } 
        return true; 
       } 
       if (candidate.getX2() < line.getX1()) 
        i.remove(); 
      } 
      candidates.add(line); 
     } 
     return false; 
    } 


    /** 
    * Returns all lines in a path. The lines are constructed such that the starting point is found 
    * on the left (or same x-coordinate) of the ending point. 
    * <p/> 
    * @param path the path 
    * @return the lines, sorted in ascending order of the x-coordinate of the starting point and 
    * ending point, respectively 
    */ 
    private static SortedSet<Line2D> getLines(PathIterator path) 
    { 
     double[] coords = new double[6]; 
     SortedSet<Line2D> result = new TreeSet<Line2D>(new Comparator<Line2D>() 
     { 
      @Override 
      public int compare(Line2D o1, Line2D o2) 
      { 
       int result = Double.compare(o1.getX1(), o2.getX1()); 
       if (result == 0) 
       { 
        // Ensure we are consistent with equals() 
        return Double.compare(o1.getX2(), o2.getX2()); 
       } 
       return result; 
      } 
     }); 
     if (path.isDone()) 
      return result; 
     int type = path.currentSegment(coords); 
     assert (type == PathIterator.SEG_MOVETO): type; 
     Point.Double startPoint = new Point.Double(coords[0], coords[1]); 
     Point.Double openPoint = startPoint; 
     path.next(); 

     while (!path.isDone()) 
     { 
      type = path.currentSegment(coords); 
      assert (type != PathIterator.SEG_CUBICTO && type != PathIterator.SEG_QUADTO): type; 
      switch (type) 
      { 
       case PathIterator.SEG_MOVETO: 
       { 
        openPoint = startPoint; 
        break; 
       } 
       case PathIterator.SEG_CLOSE: 
       { 
        coords[0] = openPoint.x; 
        coords[1] = openPoint.y; 
        break; 
       } 
      } 
      Point.Double endPoint = new Point.Double(coords[0], coords[1]); 
      if (Double.compare(startPoint.getX(), endPoint.getX()) < 0) 
       result.add(new Line2D.Double(startPoint, endPoint)); 
      else 
       result.add(new Line2D.Double(endPoint, startPoint)); 
      path.next(); 
      startPoint = endPoint; 
     } 
     return result; 
    } 
}