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;
}

这大概是一个稍微不那么优化的C代码版本，它来自于这个页面。

我的c++版本使用std::vector<std::pair<double, double>>和两个double作为x和y。逻辑应该与原始C代码完全相同，但我发现我的更容易阅读。我不能为表演说话。

bool point_in_poly(std::vector<std::pair<double, double>>& verts, double point_x, double point_y)
{
    bool in_poly = false;
    auto num_verts = verts.size();
    for (int i = 0, j = num_verts - 1; i < num_verts; j = i++) {
        double x1 = verts[i].first;
        double y1 = verts[i].second;
        double x2 = verts[j].first;
        double y2 = verts[j].second;

        if (((y1 > point_y) != (y2 > point_y)) &&
            (point_x < (x2 - x1) * (point_y - y1) / (y2 - y1) + x1))
            in_poly = !in_poly;
    }
    return in_poly;
}

原始的C代码是

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;
}

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;
}

nirg回答的Swift版本:

extension CGPoint {
    func isInsidePolygon(vertices: [CGPoint]) -> Bool {
        guard !vertices.isEmpty else { return false }
        var j = vertices.last!, c = false
        for i in vertices {
            let a = (i.y > y) != (j.y > y)
            let b = (x < (j.x - i.x) * (y - i.y) / (j.y - i.y) + i.x)
            if a && b { c = !c }
            j = i
        }
        return c
    }
}

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

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