下面是nirg给出的答案的c#版本，它来自RPI教授。请注意，使用来自RPI源代码的代码需要归属。

在顶部添加了一个边界框复选。然而，正如James Brown所指出的，主代码几乎和边界框检查本身一样快，所以边界框检查实际上会减慢整体操作，因为您正在检查的大多数点都在边界框内。所以你可以让边界框签出，或者另一种选择是预先计算多边形的边界框，如果它们不经常改变形状的话。

public bool IsPointInPolygon( Point p, Point[] polygon )
{
    double minX = polygon[ 0 ].X;
    double maxX = polygon[ 0 ].X;
    double minY = polygon[ 0 ].Y;
    double maxY = polygon[ 0 ].Y;
    for ( int i = 1 ; i < polygon.Length ; i++ )
    {
        Point q = polygon[ i ];
        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;
    }

    // https://wrf.ecse.rpi.edu/Research/Short_Notes/pnpoly.html
    bool inside = false;
    for ( int i = 0, j = polygon.Length - 1 ; 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 )
        {
            inside = !inside;
        }
    }

    return inside;
}

For graphics, I'd rather not prefer integers. Many systems use integers for UI painting (pixels are ints after all), but macOS, for example, uses float for everything. macOS only knows points and a point can translate to one pixel, but depending on monitor resolution, it might translate to something else. On retina screens half a point (0.5/0.5) is pixel. Still, I never noticed that macOS UIs are significantly slower than other UIs. After all, 3D APIs (OpenGL or Direct3D) also work with floats and modern graphics libraries very often take advantage of GPU acceleration.

现在你说速度是你最关心的，好吧，让我们追求速度。在运行任何复杂的算法之前，先做一个简单的测试。在多边形周围创建一个轴对齐的包围框。这是非常简单，快速的，已经可以节省你很多计算。这是怎么做到的呢?遍历多边形的所有点，找到X和Y的最小/最大值。

如你有点(9/1),(4/3),(2/7),(8/2),(3/6)。这意味着Xmin是2,Xmax是9,Ymin是1,Ymax是7。矩形外有两条边(2/1)和(9/7)的点不可能在多边形内。

// p is your point, p.x is the x coord, p.y is the y coord
if (p.x < Xmin || p.x > Xmax || p.y < Ymin || p.y > Ymax) {
    // Definitely not within the polygon!
}

这是对任意点运行的第一个测试。正如你所看到的，这个测试非常快，但也非常粗糙。要处理边界矩形内的点，我们需要更复杂的算法。有几种计算方法。哪种方法有效还取决于多边形是否有孔或始终是固体。以下是实体的例子(一个凸面，一个凹面):

这里有一个洞:

绿色的中间有个洞!

最简单的算法，可以处理上述三种情况，并且仍然非常快，叫做射线投射。该算法的思想非常简单:从多边形外的任何地方绘制一条虚拟光线到你的点，并计算它击中多边形一侧的频率。如果命中次数是偶数，则在多边形外，如果是奇数，则在多边形内。

圈数算法是另一种选择，它对非常接近多边形线的点更准确，但也慢得多。由于有限的浮点精度和舍入问题，光线投射可能会因为太靠近多边形一侧的点而失败，但在现实中这几乎不是问题，因为如果一个点靠近一侧，在视觉上甚至不可能让观看者识别它是否已经在内部或仍然在外部。

还记得上面的边界框吗?只需在边界框外选择一个点，并将其用作射线的起点。例如，点(Xmin - e/p.y)肯定在多边形外。

But what is e? Well, e (actually epsilon) gives the bounding box some padding. As I said, ray tracing fails if we start too close to a polygon line. Since the bounding box might equal the polygon (if the polygon is an axis aligned rectangle, the bounding box is equal to the polygon itself!), we need some padding to make this safe, that's all. How big should you choose e? Not too big. It depends on the coordinate system scale you use for drawing. If your pixel step width is 1.0, then just choose 1.0 (yet 0.1 would have worked as well)

现在我们有了光线的起始坐标和结束坐标，问题从“是多边形内的点”转移到“光线与多边形边相交的频率”。因此，我们不能像以前那样只处理多边形点，现在我们需要实际的边。一条边总是由两点来定义。

side 1: (X1/Y1)-(X2/Y2)
side 2: (X2/Y2)-(X3/Y3)
side 3: (X3/Y3)-(X4/Y4)
:

你需要从各个方面测试光线。假设射线是一个矢量，每条边都是一个矢量。光线必须恰好击中每边一次，否则就永远不会。它不可能击中同一侧两次。二维空间中的两条直线总是只相交一次，除非它们是平行的，在这种情况下，它们永远不会相交。然而，由于向量的长度是有限的，两个向量可能不平行，也永远不会相交，因为它们太短而无法相遇。

// Test the ray against all sides
int intersections = 0;
for (side = 0; side < numberOfSides; side++) {
    // Test if current side intersects with ray.
    // If yes, intersections++;
}
if ((intersections & 1) == 1) {
    // Inside of polygon
} else {
    // Outside of polygon
}

到目前为止一切顺利，但是如何检验两个向量是否相交呢?下面是一些C代码(未测试)，应该可以做到:

#define NO 0
#define YES 1
#define COLLINEAR 2

int areIntersecting(
    float v1x1, float v1y1, float v1x2, float v1y2,
    float v2x1, float v2y1, float v2x2, float v2y2
) {
    float d1, d2;
    float a1, a2, b1, b2, c1, c2;

    // Convert vector 1 to a line (line 1) of infinite length.
    // We want the line in linear equation standard form: A*x + B*y + C = 0
    // See: http://en.wikipedia.org/wiki/Linear_equation
    a1 = v1y2 - v1y1;
    b1 = v1x1 - v1x2;
    c1 = (v1x2 * v1y1) - (v1x1 * v1y2);

    // Every point (x,y), that solves the equation above, is on the line,
    // every point that does not solve it, is not. The equation will have a
    // positive result if it is on one side of the line and a negative one 
    // if is on the other side of it. We insert (x1,y1) and (x2,y2) of vector
    // 2 into the equation above.
    d1 = (a1 * v2x1) + (b1 * v2y1) + c1;
    d2 = (a1 * v2x2) + (b1 * v2y2) + c1;

    // If d1 and d2 both have the same sign, they are both on the same side
    // of our line 1 and in that case no intersection is possible. Careful, 
    // 0 is a special case, that's why we don't test ">=" and "<=", 
    // but "<" and ">".
    if (d1 > 0 && d2 > 0) return NO;
    if (d1 < 0 && d2 < 0) return NO;

    // The fact that vector 2 intersected the infinite line 1 above doesn't 
    // mean it also intersects the vector 1. Vector 1 is only a subset of that
    // infinite line 1, so it may have intersected that line before the vector
    // started or after it ended. To know for sure, we have to repeat the
    // the same test the other way round. We start by calculating the 
    // infinite line 2 in linear equation standard form.
    a2 = v2y2 - v2y1;
    b2 = v2x1 - v2x2;
    c2 = (v2x2 * v2y1) - (v2x1 * v2y2);

    // Calculate d1 and d2 again, this time using points of vector 1.
    d1 = (a2 * v1x1) + (b2 * v1y1) + c2;
    d2 = (a2 * v1x2) + (b2 * v1y2) + c2;

    // Again, if both have the same sign (and neither one is 0),
    // no intersection is possible.
    if (d1 > 0 && d2 > 0) return NO;
    if (d1 < 0 && d2 < 0) return NO;

    // If we get here, only two possibilities are left. Either the two
    // vectors intersect in exactly one point or they are collinear, which
    // means they intersect in any number of points from zero to infinite.
    if ((a1 * b2) - (a2 * b1) == 0.0f) return COLLINEAR;

    // If they are not collinear, they must intersect in exactly one point.
    return YES;
}

输入值是向量1 (v1x1/v1y1和v1x2/v1y2)和向量2 (v2x1/v2y1和v2x2/v2y2)的两个端点。有2个向量，4个点，8个坐标。YES和NO很清楚。“是”增加了交叉路口，“否”什么都不做。

What about COLLINEAR? It means both vectors lie on the same infinite line, depending on position and length, they don't intersect at all or they intersect in an endless number of points. I'm not absolutely sure how to handle this case, I would not count it as intersection either way. Well, this case is rather rare in practice anyway because of floating point rounding errors; better code would probably not test for == 0.0f but instead for something like < epsilon, where epsilon is a rather small number.

如果你需要测试更多的点，你当然可以通过在内存中保留多边形边的线性方程标准形式来加快整个过程，这样你就不必每次都重新计算这些点。这将在每次测试中为您节省两次浮点乘法和三次浮点减法，以换取在内存中为每个多边形边存储三个浮点值。这是一个典型的内存与计算时间的权衡。

Last but not least: If you may use 3D hardware to solve the problem, there is an interesting alternative. Just let the GPU do all the work for you. Create a painting surface that is off screen. Fill it completely with the color black. Now let OpenGL or Direct3D paint your polygon (or even all of your polygons if you just want to test if the point is within any of them, but you don't care for which one) and fill the polygon(s) with a different color, e.g. white. To check if a point is within the polygon, get the color of this point from the drawing surface. This is just a O(1) memory fetch.

Of course this method is only usable if your drawing surface doesn't have to be huge. If it cannot fit into the GPU memory, this method is slower than doing it on the CPU. If it would have to be huge and your GPU supports modern shaders, you can still use the GPU by implementing the ray casting shown above as a GPU shader, which absolutely is possible. For a larger number of polygons or a large number of points to test, this will pay off, consider some GPUs will be able to test 64 to 256 points in parallel. Note however that transferring data from CPU to GPU and back is always expensive, so for just testing a couple of points against a couple of simple polygons, where either the points or the polygons are dynamic and will change frequently, a GPU approach will rarely pay off.

如果你正在寻找一个java脚本库，有一个javascript谷歌maps v3扩展的Polygon类，以检测是否有一个点驻留在它里面。

var polygon = new google.maps.Polygon([], "#000000", 1, 1, "#336699", 0.3);
var isWithinPolygon = polygon.containsLatLng(40, -90);

谷歌扩展Github

这个问题很有趣。我有另一个可行的想法，不同于这篇文章的其他答案。其原理是利用角度之和来判断目标是在内部还是外部。也就是圈数。

设x为目标点。让数组[0,1，....N]是该区域的所有点。用一条线将目标点与每一个边界点连接起来。如果目标点在这个区域内。所有角的和是360度。如果不是，角度将小于360度。

参考这张图来对这个概念有一个基本的了解:

我的算法假设顺时针是正方向。这是一个潜在的输入:

[[-122.402015, 48.225216], [-117.032049, 48.999931], [-116.919132, 45.995175], [-124.079107, 46.267259], [-124.717175, 48.377557], [-122.92315, 47.047963], [-122.402015, 48.225216]]

下面是实现这个想法的python代码:

def isInside(self, border, target):
degree = 0
for i in range(len(border) - 1):
    a = border[i]
    b = border[i + 1]

    # calculate distance of vector
    A = getDistance(a[0], a[1], b[0], b[1]);
    B = getDistance(target[0], target[1], a[0], a[1])
    C = getDistance(target[0], target[1], b[0], b[1])

    # calculate direction of vector
    ta_x = a[0] - target[0]
    ta_y = a[1] - target[1]
    tb_x = b[0] - target[0]
    tb_y = b[1] - target[1]

    cross = tb_y * ta_x - tb_x * ta_y
    clockwise = cross < 0

    # calculate sum of angles
    if(clockwise):
        degree = degree + math.degrees(math.acos((B * B + C * C - A * A) / (2.0 * B * C)))
    else:
        degree = degree - math.degrees(math.acos((B * B + C * C - A * A) / (2.0 * B * C)))

if(abs(round(degree) - 360) <= 3):
    return True
return False

下面是Rust版本的@nirg答案(Philipp Lenssen javascript版本) 我给出这个答案是因为我从这个网站得到了很多帮助，我翻译javascript版本rust作为一个练习，希望可以帮助一些人，最后一个原因是，在我的工作中，我会把这段代码翻译成一个wasm，以提高我的画布的性能，这是一个开始。我的英语很差……,请原谅我 ｀

pub struct Point {
    x: f32,
    y: f32,
}
pub fn point_is_in_poly(pt: Point, polygon: &Vec<Point>) -> bool {
    let mut is_inside = false;

    let max_x = polygon.iter().map(|pt| pt.x).reduce(f32::max).unwrap();
    let min_x = polygon.iter().map(|pt| pt.x).reduce(f32::min).unwrap();
    let max_y = polygon.iter().map(|pt| pt.y).reduce(f32::max).unwrap();
    let min_y = polygon.iter().map(|pt| pt.y).reduce(f32::min).unwrap();

    if pt.x < min_x || pt.x > max_x || pt.y < min_y || pt.y > max_y {
        return is_inside;
    }

    let len = polygon.len();
    let mut j = len - 1;

    for i in 0..len {
        let y_i_value = polygon[i].y > pt.y;
        let y_j_value = polygon[j].y > pt.y;
        let last_check = (polygon[j].x - polygon[i].x) * (pt.y - polygon[i].y)
            / (polygon[j].y - polygon[i].y)
            + polygon[i].x;
        if y_i_value != y_j_value && pt.x < last_check {
            is_inside = !is_inside;
        }
        j = i;
    }
    is_inside
}


let pt = Point {
    x: 1266.753,
    y: 97.655,
};
let polygon = vec![
    Point {
        x: 725.278,
        y: 203.586,
    },
    Point {
        x: 486.831,
        y: 441.931,
    },
    Point {
        x: 905.77,
        y: 445.241,
    },
    Point {
        x: 1026.649,
        y: 201.931,
    },
];
let pt1 = Point {
    x: 725.278,
    y: 203.586,
};
let pt2 = Point {
    x: 872.652,
    y: 321.103,
};
println!("{}", point_is_in_poly(pt, &polygon));// false
println!("{}", point_is_in_poly(pt1, &polygon)); // true
println!("{}", point_is_in_poly(pt2, &polygon));// true

`

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