David Segond's answer is pretty much the standard general answer, and Richard T's is the most common optimization, though therre are some others. Other strong optimizations are based on less general solutions. For example if you are going to check the same polygon with lots of points, triangulating the polygon can speed things up hugely as there are a number of very fast TIN searching algorithms. Another is if the polygon and points are on a limited plane at low resolution, say a screen display, you can paint the polygon onto a memory mapped display buffer in a given colour, and check the color of a given pixel to see if it lies in the polygons.

像许多优化一样，这些优化是基于特定情况而不是一般情况，并且基于摊销时间而不是单次使用产生效益。

在这个领域工作，我发现约瑟夫·奥鲁克斯的《计算几何》在C' ISBN 0-521-44034-3是一个很大的帮助。

我知道这是旧的，但这里是一个在Cocoa实现的光线投射算法，如果有人感兴趣的话。不确定这是最有效的方法，但它可能会帮助别人。

- (BOOL)shape:(NSBezierPath *)path containsPoint:(NSPoint)point
{
    NSBezierPath *currentPath = [path bezierPathByFlatteningPath];
    BOOL result;
    float aggregateX = 0; //I use these to calculate the centroid of the shape
    float aggregateY = 0;
    NSPoint firstPoint[1];
    [currentPath elementAtIndex:0 associatedPoints:firstPoint];
    float olderX = firstPoint[0].x;
    float olderY = firstPoint[0].y;
    NSPoint interPoint;
    int noOfIntersections = 0;

    for (int n = 0; n < [currentPath elementCount]; n++) {
        NSPoint points[1];
        [currentPath elementAtIndex:n associatedPoints:points];
        aggregateX += points[0].x;
        aggregateY += points[0].y;
    }

    for (int n = 0; n < [currentPath elementCount]; n++) {
        NSPoint points[1];

        [currentPath elementAtIndex:n associatedPoints:points];
        //line equations in Ax + By = C form
        float _A_FOO = (aggregateY/[currentPath elementCount]) - point.y;  
        float _B_FOO = point.x - (aggregateX/[currentPath elementCount]);
        float _C_FOO = (_A_FOO * point.x) + (_B_FOO * point.y);

        float _A_BAR = olderY - points[0].y;
        float _B_BAR = points[0].x - olderX;
        float _C_BAR = (_A_BAR * olderX) + (_B_BAR * olderY);

        float det = (_A_FOO * _B_BAR) - (_A_BAR * _B_FOO);
        if (det != 0) {
            //intersection points with the edges
            float xIntersectionPoint = ((_B_BAR * _C_FOO) - (_B_FOO * _C_BAR)) / det;
            float yIntersectionPoint = ((_A_FOO * _C_BAR) - (_A_BAR * _C_FOO)) / det;
            interPoint = NSMakePoint(xIntersectionPoint, yIntersectionPoint);
            if (olderX <= points[0].x) {
                //doesn't matter in which direction the ray goes, so I send it right-ward.
                if ((interPoint.x >= olderX && interPoint.x <= points[0].x) && (interPoint.x > point.x)) {  
                    noOfIntersections++;
                }
            } else {
                if ((interPoint.x >= points[0].x && interPoint.x <= olderX) && (interPoint.x > point.x)) {
                     noOfIntersections++;
                } 
            }
        }
        olderX = points[0].x;
        olderY = points[0].y;
    }
    if (noOfIntersections % 2 == 0) {
        result = FALSE;
    } else {
        result = TRUE;
    }
    return result;
}

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

Scala版本的解决方案由nirg(假设边界矩形预检查是单独完成的):

def inside(p: Point, polygon: Array[Point], bounds: Bounds): Boolean = {

  val length = polygon.length

  @tailrec
  def oddIntersections(i: Int, j: Int, tracker: Boolean): Boolean = {
    if (i == length)
      tracker
    else {
      val intersects = (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
      oddIntersections(i + 1, i, if (intersects) !tracker else tracker)
    }
  }

  oddIntersections(0, length - 1, tracker = false)
}

简单的解决方案是将多边形划分为三角形，并按这里解释的那样对三角形进行测试

如果你的多边形是凸多边形，可能有更好的方法。把这个多边形看作是无限条线的集合。每一行将空间一分为二。对于每一个点，很容易判断它是在直线的一边还是另一边。如果一个点在所有直线的同一侧，那么它在多边形内。

nirg的c#版本的答案在这里:我只分享代码。这可能会节省一些时间。

public static bool IsPointInPolygon(IList<Point> polygon, Point testPoint) {
            bool result = false;
            int j = polygon.Count() - 1;
            for (int i = 0; i < polygon.Count(); i++) {
                if (polygon[i].Y < testPoint.Y && polygon[j].Y >= testPoint.Y || polygon[j].Y < testPoint.Y && polygon[i].Y >= testPoint.Y) {
                    if (polygon[i].X + (testPoint.Y - polygon[i].Y) / (polygon[j].Y - polygon[i].Y) * (polygon[j].X - polygon[i].X) < testPoint.X) {
                        result = !result;
                    }
                }
                j = i;
            }
            return result;
        }