如果你正在使用谷歌Map SDK，想要检查一个点是否在一个多边形内，你可以尝试使用GMSGeometryContainsLocation。效果很好!!它是这样运作的，

if GMSGeometryContainsLocation(point, polygon, true) {
    print("Inside this polygon.")
} else {
    print("outside this polygon")
}

这里是参考资料:https://developers.google.com/maps/documentation/ios-sdk/reference/group___geometry_utils#gaba958d3776d49213404af249419d0ffd

VBA版本:

注意:请记住，如果你的多边形是地图中的一个区域，纬度/经度是Y/X值，而不是X/Y(纬度= Y，经度= X)，因为从我的理解来看，这是历史含义，因为经度不是一个测量值。

类模块:CPoint

Private pXValue As Double
Private pYValue As Double

'''''X Value Property'''''

Public Property Get X() As Double
    X = pXValue
End Property

Public Property Let X(Value As Double)
    pXValue = Value
End Property

'''''Y Value Property'''''

Public Property Get Y() As Double
    Y = pYValue
End Property

Public Property Let Y(Value As Double)
    pYValue = Value
End Property

模块:

Public Function isPointInPolygon(p As CPoint, polygon() As CPoint) As Boolean

    Dim i As Integer
    Dim j As Integer
    Dim q As Object
    Dim minX As Double
    Dim maxX As Double
    Dim minY As Double
    Dim maxY As Double
    minX = polygon(0).X
    maxX = polygon(0).X
    minY = polygon(0).Y
    maxY = polygon(0).Y

    For i = 1 To UBound(polygon)
        Set q = polygon(i)
        minX = vbMin(q.X, minX)
        maxX = vbMax(q.X, maxX)
        minY = vbMin(q.Y, minY)
        maxY = vbMax(q.Y, maxY)
    Next i

    If p.X < minX Or p.X > maxX Or p.Y < minY Or p.Y > maxY Then
        isPointInPolygon = False
        Exit Function
    End If


    ' SOURCE: http://www.ecse.rpi.edu/Homepages/wrf/Research/Short_Notes/pnpoly.html

    isPointInPolygon = False
    i = 0
    j = UBound(polygon)

    Do While i < UBound(polygon) + 1
        If (polygon(i).Y > p.Y) Then
            If (polygon(j).Y < p.Y) Then
                If p.X < (polygon(j).X - polygon(i).X) * (p.Y - polygon(i).Y) / (polygon(j).Y - polygon(i).Y) + polygon(i).X Then
                    isPointInPolygon = True
                    Exit Function
                End If
            End If
        ElseIf (polygon(i).Y < p.Y) Then
            If (polygon(j).Y > p.Y) Then
                If p.X < (polygon(j).X - polygon(i).X) * (p.Y - polygon(i).Y) / (polygon(j).Y - polygon(i).Y) + polygon(i).X Then
                    isPointInPolygon = True
                    Exit Function
                End If
            End If
        End If
        j = i
        i = i + 1
    Loop   
End Function

Function vbMax(n1, n2) As Double
    vbMax = IIf(n1 > n2, n1, n2)
End Function

Function vbMin(n1, n2) As Double
    vbMin = IIf(n1 > n2, n2, n1)
End Function


Sub TestPointInPolygon()

    Dim i As Integer
    Dim InPolygon As Boolean

'   MARKER Object
    Dim p As CPoint
    Set p = New CPoint
    p.X = <ENTER X VALUE HERE>
    p.Y = <ENTER Y VALUE HERE>

'   POLYGON OBJECT
    Dim polygon() As CPoint
    ReDim polygon(<ENTER VALUE HERE>) 'Amount of vertices in polygon - 1
    For i = 0 To <ENTER VALUE HERE> 'Same value as above
       Set polygon(i) = New CPoint
       polygon(i).X = <ASSIGN X VALUE HERE> 'Source a list of values that can be looped through
       polgyon(i).Y = <ASSIGN Y VALUE HERE> 'Source a list of values that can be looped through
    Next i

    InPolygon = isPointInPolygon(p, polygon)
    MsgBox InPolygon

End Sub

以下是M. Katz基于Nirg方法的答案的JavaScript变体:

function pointIsInPoly(p, polygon) {
    var isInside = false;
    var minX = polygon[0].x, maxX = polygon[0].x;
    var minY = polygon[0].y, maxY = polygon[0].y;
    for (var n = 1; n < polygon.length; n++) {
        var q = polygon[n];
        minX = Math.min(q.x, minX);
        maxX = Math.max(q.x, maxX);
        minY = Math.min(q.y, minY);
        maxY = Math.max(q.y, maxY);
    }

    if (p.x < minX || p.x > maxX || p.y < minY || p.y > maxY) {
        return false;
    }

    var i = 0, j = polygon.length - 1;
    for (i, j; i < polygon.length; j = i++) {
        if ( (polygon[i].y > p.y) != (polygon[j].y > p.y) &&
                p.x < (polygon[j].x - polygon[i].x) * (p.y - polygon[i].y) / (polygon[j].y - polygon[i].y) + polygon[i].x ) {
            isInside = !isInside;
        }
    }

    return isInside;
}

如果你正在使用谷歌Map SDK，想要检查一个点是否在一个多边形内，你可以尝试使用GMSGeometryContainsLocation。效果很好!!它是这样运作的，

if GMSGeometryContainsLocation(point, polygon, true) {
    print("Inside this polygon.")
} else {
    print("outside this polygon")
}

这里是参考资料:https://developers.google.com/maps/documentation/ios-sdk/reference/group___geometry_utils#gaba958d3776d49213404af249419d0ffd

Java版本:

public class Geocode {
    private float latitude;
    private float longitude;

    public Geocode() {
    }

    public Geocode(float latitude, float longitude) {
        this.latitude = latitude;
        this.longitude = longitude;
    }

    public float getLatitude() {
        return latitude;
    }

    public void setLatitude(float latitude) {
        this.latitude = latitude;
    }

    public float getLongitude() {
        return longitude;
    }

    public void setLongitude(float longitude) {
        this.longitude = longitude;
    }
}

public class GeoPolygon {
    private ArrayList<Geocode> points;

    public GeoPolygon() {
        this.points = new ArrayList<Geocode>();
    }

    public GeoPolygon(ArrayList<Geocode> points) {
        this.points = points;
    }

    public GeoPolygon add(Geocode geo) {
        points.add(geo);
        return this;
    }

    public boolean inside(Geocode geo) {
        int i, j;
        boolean c = false;
        for (i = 0, j = points.size() - 1; i < points.size(); j = i++) {
            if (((points.get(i).getLongitude() > geo.getLongitude()) != (points.get(j).getLongitude() > geo.getLongitude())) &&
                    (geo.getLatitude() < (points.get(j).getLatitude() - points.get(i).getLatitude()) * (geo.getLongitude() - points.get(i).getLongitude()) / (points.get(j).getLongitude() - points.get(i).getLongitude()) + points.get(i).getLatitude()))
                c = !c;
        }
        return c;
    }

}

我认为下面这段代码是最好的解决方案(从这里开始):

int pnpoly(int nvert, float *vertx, float *verty, float testx, float testy)
{
  int i, j, c = 0;
  for (i = 0, j = nvert-1; i < nvert; j = i++) {
    if ( ((verty[i]>testy) != (verty[j]>testy)) &&
     (testx < (vertx[j]-vertx[i]) * (testy-verty[i]) / (verty[j]-verty[i]) + vertx[i]) )
       c = !c;
  }
  return c;
}

参数

nvert:多边形中的顶点数。是否在末端重复第一个顶点在上面的文章中已经讨论过了。 vertx, verty:包含多边形顶点的x坐标和y坐标的数组。 testx, testy:测试点的X坐标和y坐标。

它既简短又高效，适用于凸多边形和凹多边形。如前所述，您应该首先检查边界矩形，并单独处理多边形孔。

这背后的想法很简单。作者描述如下:

我从测试点水平运行一条半无限射线(增加x，固定y)，并计算它穿过多少条边。在每个十字路口，光线在内部和外部之间切换。这叫做乔丹曲线定理。

当水平射线穿过任意一条边时，变量c从0变为1，从1变为0。基本上它记录了交叉边的数量是偶数还是奇数。0表示偶数，1表示奇数。