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

Obj-C版本nirg的答案与样本方法测试点。Nirg的回答对我很有效。

- (BOOL)isPointInPolygon:(NSArray *)vertices point:(CGPoint)test {
    NSUInteger nvert = [vertices count];
    NSInteger i, j, c = 0;
    CGPoint verti, vertj;

    for (i = 0, j = nvert-1; i < nvert; j = i++) {
        verti = [(NSValue *)[vertices objectAtIndex:i] CGPointValue];
        vertj = [(NSValue *)[vertices objectAtIndex:j] CGPointValue];
        if (( (verti.y > test.y) != (vertj.y > test.y) ) &&
        ( test.x < ( vertj.x - verti.x ) * ( test.y - verti.y ) / ( vertj.y - verti.y ) + verti.x) )
            c = !c;
    }

    return (c ? YES : NO);
}

- (void)testPoint {

    NSArray *polygonVertices = [NSArray arrayWithObjects:
        [NSValue valueWithCGPoint:CGPointMake(13.5, 41.5)],
        [NSValue valueWithCGPoint:CGPointMake(42.5, 56.5)],
        [NSValue valueWithCGPoint:CGPointMake(39.5, 69.5)],
        [NSValue valueWithCGPoint:CGPointMake(42.5, 84.5)],
        [NSValue valueWithCGPoint:CGPointMake(13.5, 100.0)],
        [NSValue valueWithCGPoint:CGPointMake(6.0, 70.5)],
        nil
    ];

    CGPoint tappedPoint = CGPointMake(23.0, 70.0);

    if ([self isPointInPolygon:polygonVertices point:tappedPoint]) {
        NSLog(@"YES");
    } else {
        NSLog(@"NO");
    }
}

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。

我已经做了nirg的c++代码的Python实现:

输入

Bounding_points:组成多边形的节点。 Bounding_box_positions:筛选的候选点。(在我从边界框创建的实现中。 (输入为元组列表，格式为:[(xcord, ycord)，…])

返回

多边形内的所有点。

def polygon_ray_casting(self, bounding_points, bounding_box_positions):
    # Arrays containing the x- and y-coordinates of the polygon's vertices.
    vertx = [point[0] for point in bounding_points]
    verty = [point[1] for point in bounding_points]
    # Number of vertices in the polygon
    nvert = len(bounding_points)
    # Points that are inside
    points_inside = []

    # For every candidate position within the bounding box
    for idx, pos in enumerate(bounding_box_positions):
        testx, testy = (pos[0], pos[1])
        c = 0
        for i in range(0, nvert):
            j = i - 1 if i != 0 else nvert - 1
            if( ((verty[i] > testy ) != (verty[j] > testy))   and
                    (testx < (vertx[j] - vertx[i]) * (testy - verty[i]) / (verty[j] - verty[i]) + vertx[i]) ):
                c += 1
        # If odd, that means that we are inside the polygon
        if c % 2 == 1: 
            points_inside.append(pos)


    return points_inside

同样，这个想法也是从这里得来的

当使用qt (qt 4.3+)时，可以使用QPolygon的函数containsPoint

from typing import Iterable

def pnpoly(verts, x, y):
    #check if x and/or y is iterable
    xit, yit = isinstance(x, Iterable), isinstance(y, Iterable)
    #if not iterable, make an iterable of length 1
    X = x if xit else (x, )
    Y = y if yit else (y, )
    #store verts length as a range to juggle j
    r = range(len(verts))
    #final results if x or y is iterable
    results = []
    #traverse x and y coordinates
    for xp in X:
        for yp in Y:
            c = 0 #reset c at every new position
            for i in r:
                j = r[i-1] #set j to position before i
                #store a few arguments to shorten the if statement
                yneq       = (verts[i][1] > yp) != (verts[j][1] > yp)
                xofs, yofs = (verts[j][0] - verts[i][0]), (verts[j][1] - verts[i][1])
                #if we have crossed a line, increment c
                if (yneq and (xp < xofs * (yp - verts[i][1]) / yofs + verts[i][0])):
                    c += 1
            #if c is odd store the coordinates        
            if c%2:
                results.append((xp, yp))
    #return either coordinates or a bool, depending if x or y was an iterable
    return results if (xit or yit) else bool(c%2)

这个python版本是通用的。您可以为True/False结果输入单个x和单个y值，也可以使用x和y的范围来遍历整个点网格。如果使用范围，则返回所有True点的x/y对列表。vertices参数需要一个由x/y对组成的二维Iterable，例如:[(x1,y1)， (x2,y2)，…]

使用示例:

vertices = [(25,25), (75,25), (75,75), (25,75)]
pnpoly(vertices, 50, 50) #True
pnpoly(vertices, range(100), range(100)) #[(25,25), (25,26), (25,27), ...]

实际上，这些都可以。

pnpoly(vertices, 50, range(100)) #check 0 to 99 y at x of 50
pnpoly(vertices, range(100), 50) #check 0 to 99 x at y of 50