if(myPath.contains(x,y)
到目前为止，我的选项是：
我前不久也遇到了同样的问题，在搜索后，我发现这是最好的解决方案。
Java有一个带有contains()方法的Polygon类，可以使事情变得非常简单。不幸的是，java.awt.Polygon类在Android上不受支持。但是，我找到了一个人编写的等效类。
我不认为您可以从AndroidPath类中获取构成路径的单个点，因此您将不得不以不同的方式存储数据。
该类使用交叉数算法来确定给定点是否在给定点列表内部。
/**
* Minimum Polygon class for Android.
*/
public class Polygon
{
// Polygon coodinates.
private int[] polyY, polyX;
// Number of sides in the polygon.
private int polySides;
/**
* Default constructor.
* @param px Polygon y coods.
* @param py Polygon x coods.
* @param ps Polygon sides count.
*/
public Polygon( int[] px, int[] py, int ps )
{
polyX = px;
polyY = py;
polySides = ps;
}
/**
* Checks if the Polygon contains a point.
* @see "http://alienryderflex.com/polygon/"
* @param x Point horizontal pos.
* @param y Point vertical pos.
* @return Point is in Poly flag.
*/
public boolean contains( int x, int y )
{
boolean oddTransitions = false;
for( int i = 0, j = polySides -1; i < polySides; j = i++ )
{
if( ( polyY[ i ] < y && polyY[ j ] >= y ) || ( polyY[ j ] < y && polyY[ i ] >= y ) )
{
if( polyX[ i ] + ( y - polyY[ i ] ) / ( polyY[ j ] - polyY[ i ] ) * ( polyX[ j ] - polyX[ i ] ) < x )
{
oddTransitions = !oddTransitions;
}
}
}
return oddTransitions;
}
}
/**
* Minimum Polygon class for Android.
*/
public class Polygon
{
// Polygon coodinates.
private int[] polyY, polyX;
// Number of sides in the polygon.
private int polySides;
/**
* Default constructor.
* @param px Polygon y coods.
* @param py Polygon x coods.
* @param ps Polygon sides count.
*/
public Polygon( int[] px, int[] py, int ps )
{
polyX = px;
polyY = py;
polySides = ps;
}
/**
* Checks if the Polygon contains a point.
* @see "http://alienryderflex.com/polygon/"
* @param x Point horizontal pos.
* @param y Point vertical pos.
* @return Point is in Poly flag.
*/
public boolean contains( int x, int y )
{
boolean c = false;
int i, j = 0;
for (i = 0, j = polySides - 1; i < polySides; j = i++) {
if (((polyY[i] > y) != (polyY[j] > y))
&& (x < (polyX[j] - polyX[i]) * (y - polyY[i]) / (polyY[j] - polyY[i]) + polyX[i]))
c = !c;
}
return c;
}
}
为了完整起见，在这里我想做几个注释：
从API 19开始，路径有一个交集操作。您可以在测试点周围创建一个非常小的正方形路径，将其与路径相交，并查看结果是否为空。
您可以将路径转换为区域并执行包含()操作。但是，区域使用整数坐标，并且我认为它们使用变换后的（像素）坐标，因此您必须处理这些内容。我还怀疑转换过程计算密集。
Hans发布的边界穿越算法很好且快速，但是您必须非常小心，以避免某些角落情况，例如光线直接通过顶点、相交水平边缘或舍入误差成为问题时，这总是存在的。
绕数方法非常可靠，但涉及大量三角函数计算，计算成本高。
Dan Sunday的这篇论文提供了一种混合算法，与绕数一样准确，但计算简单，类似于射线投射算法。它的优雅之处让我惊叹不已。
这是我最近用Java编写的一些代码，用于处理由线段和圆弧组成的路径。（还有圆，但那些是自成一体的完整路径，所以这是一种退化情况。）
package org.efalk.util;
/**
* Utility: determine if a point is inside a path.
*/
public class PathUtil {
static final double RAD = (Math.PI/180.);
static final double DEG = (180./Math.PI);
protected static final int LINE = 0;
protected static final int ARC = 1;
protected static final int CIRCLE = 2;
/**
* Used to cache the contents of a path for pick testing. For a
* line segment, x0,y0,x1,y1 are the endpoints of the line. For
* a circle (ellipse, actually), x0,y0,x1,y1 are the bounding box
* of the circle (this is how Android and X11 like to represent
* circles). For an arc, x0,y0,x1,y1 are the bounding box, a1 is
* the start angle (degrees CCW from the +X direction) and a1 is
* the sweep angle (degrees CCW).
*/
public static class PathElement {
public int type;
public float x0,y0,x1,y1; // Endpoints or bounding box
public float a0,a1; // Arcs and circles
}
/**
* Determine if the given point is inside the given path.
*/
public static boolean inside(float x, float y, PathElement[] path) {
// Based on algorithm by Dan Sunday, but allows for arc segments too.
// http://geomalgorithms.com/a03-_inclusion.html
int wn = 0;
// loop through all edges of the polygon
// An upward crossing requires y0 <= y and y1 > y
// A downward crossing requires y0 > y and y1 <= y
for (PathElement pe : path) {
switch (pe.type) {
case LINE:
if (pe.x0 < x && pe.x1 < x) // left
break;
if (pe.y0 <= y) { // start y <= P.y
if (pe.y1 > y) { // an upward crossing
if (isLeft(pe, x, y) > 0) // P left of edge
++wn; // have a valid up intersect
}
}
else { // start y > P.y
if (pe.y1 <= y) { // a downward crossing
if (isLeft(pe, x, y) < 0) // P right of edge
--wn; // have a valid down intersect
}
}
break;
case ARC:
wn += arcCrossing(pe, x, y);
break;
case CIRCLE:
// This should be the only element in the path, so test it
// and get out.
float rx = (pe.x1-pe.x0)/2;
float ry = (pe.y1-pe.y0)/2;
float xc = (pe.x1+pe.x0)/2;
float yc = (pe.y1+pe.y0)/2;
return (x-xc)*(x-xc)/rx*rx + (y-yc)*(y-yc)/ry*ry <= 1;
}
}
return wn != 0;
}
/**
* Return >0 if p is left of line p0-p1; <0 if to the right; 0 if
* on the line.
*/
private static float
isLeft(float x0, float y0, float x1, float y1, float x, float y)
{
return (x1 - x0) * (y - y0) - (x - x0) * (y1 - y0);
}
private static float isLeft(PathElement pe, float x, float y) {
return isLeft(pe.x0,pe.y0, pe.x1,pe.y1, x,y);
}
/**
* Determine if an arc segment crosses the test ray up or down, or not
* at all.
* @return winding number increment:
* +1 upward crossing
* 0 no crossing
* -1 downward crossing
*/
private static int arcCrossing(PathElement pe, float x, float y) {
// Look for trivial reject cases first.
if (pe.x1 < x || pe.y1 < y || pe.y0 > y) return 0;
// Find the intersection of the test ray with the arc. This consists
// of finding the intersection(s) of the line with the ellipse that
// contains the arc, then determining if the intersection(s)
// are within the limits of the arc.
// Since we're mostly concerned with whether or not there *is* an
// intersection, we have several opportunities to punt.
// An upward crossing requires y0 <= y and y1 > y
// A downward crossing requires y0 > y and y1 <= y
float rx = (pe.x1-pe.x0)/2;
float ry = (pe.y1-pe.y0)/2;
float xc = (pe.x1+pe.x0)/2;
float yc = (pe.y1+pe.y0)/2;
if (rx == 0 || ry == 0) return 0;
if (rx < 0) rx = -rx;
if (ry < 0) ry = -ry;
// We start by transforming everything so the ellipse is the unit
// circle; this simplifies the math.
x -= xc;
y -= yc;
if (x > rx || y > ry || y < -ry) return 0;
x /= rx;
y /= ry;
// Now find the points of intersection. This is simplified by the
// fact that our line is horizontal. Also, by the time we get here,
// we know there *is* an intersection.
// The equation for the circle is x²+y² = 1. We have y, so solve
// for x = ±sqrt(1 - y²)
double x0 = 1 - y*y;
if (x0 <= 0) return 0;
x0 = Math.sqrt(x0);
// We only care about intersections to the right of x, so
// that's another opportunity to punt. For a CCW arc, The right
// intersection is an upward crossing and the left intersection
// is a downward crossing. The reverse is true for a CW arc.
if (x > x0) return 0;
int wn = arcXing1(x0,y, pe.a0, pe.a1);
if (x < -x0) wn -= arcXing1(-x0,y, pe.a0, pe.a1);
return wn;
}
/**
* Return the winding number of the point x,y on the unit circle
* which passes through the arc segment defined by a0,a1.
*/
private static int arcXing1(double x, float y, float a0, float a1) {
double a = Math.atan2(y,x) * DEG;
if (a < 0) a += 360;
if (a1 > 0) { // CCW
if (a < a0) a += 360;
return a0 + a1 > a ? 1 : 0;
} else { // CW
if (a0 < a) a0 += 360;
return a0 + a1 <= a ? -1 : 0;
}
}
}
import PathUtil;
import PathUtil.PathElement;
/**
* This class represents a single geographic area defined by a
* circle or a list of line segments and arcs.
*/
public class Area {
public float lat0, lon0, lat1, lon1; // bounds
Path path = null;
PathElement[] pathList;
/**
* Return true if this point is inside the area bounds. This is
* used to confirm touch events and may be computationally expensive.
*/
public boolean pointInBounds(float lat, float lon) {
if (lat < lat0 || lat > lat1 || lon < lon0 || lon > lon1)
return false;
return PathUtil.inside(lon, lat, pathList);
}
static void loadBounds() {
int n = number_of_elements_in_input;
path = new Path();
pathList = new PathElement[n];
for (Element element : elements_in_input) {
PathElement pe = new PathElement();
pathList[i] = pe;
pe.type = element.type;
switch (element.type) {
case LINE: // Line segment
pe.x0 = element.x0;
pe.y0 = element.y0;
pe.x1 = element.x1;
pe.y1 = element.y1;
// Add to path, not shown here
break;
case ARC: // Arc segment
pe.x0 = element.xmin; // Bounds of arc ellipse
pe.y0 = element.ymin;
pe.x1 = element.xmax;
pe.y1 = element.ymax;
pe.a0 = a0; pe.a1 = a1;
break;
case CIRCLE: // Circle; hopefully the only entry here
pe.x0 = element.xmin; // Bounds of ellipse
pe.y0 = element.ymin;
pe.x1 = element.xmax;
pe.y1 = element.ymax;
// Add to path, not shown here
break;
}
}
path.close();
}