有一些恼人的边条件，即一个点恰好在两个相邻三角形的公共边上。这个点不可能在两个三角形中，也不可能不在两个三角形中。你需要一种任意但一致的方式来分配点。例如，画一条横线穿过这个点。如果这条直线与三角形的另一边在右侧相交，则该点被视为在三角形内。如果交点在左边，则该点在外面。

如果该点所在的直线是水平的，则使用above/below。

如果该点位于多个三角形的公共顶点上，则使用该点与中心点形成的角最小的三角形。

更有趣的是:三个点可以在一条直线上(零度)，例如(0,0)-(0,10)-(0,5)。在三角剖分算法中，“耳朵”(0,10)必须被切掉，生成的“三角形”是直线的退化情况。

通过使用重心坐标的解析解(由Andreas Brinck指出)和:

不是把乘法分布在括号里的项上 通过存储相同的项来避免多次计算 还原比较(如coproc和Thomas Eding所指出的)

可以最小化“昂贵”操作的数量:

function ptInTriangle(p, p0, p1, p2) {
    var dX = p.x-p2.x;
    var dY = p.y-p2.y;
    var dX21 = p2.x-p1.x;
    var dY12 = p1.y-p2.y;
    var D = dY12*(p0.x-p2.x) + dX21*(p0.y-p2.y);
    var s = dY12*dX + dX21*dY;
    var t = (p2.y-p0.y)*dX + (p0.x-p2.x)*dY;
    if (D<0) return s<=0 && t<=0 && s+t>=D;
    return s>=0 && t>=0 && s+t<=D;
}

代码可以粘贴在Perro Azul jsfiddle中，或者通过点击下面的“运行代码片段”来尝试

var ctx = $("canvas")[0].getContext("2d"); var W = 500; var H = 500; var point = { x: W / 2, y: H / 2 }; var triangle = randomTriangle(); $("canvas").click(function(evt) { point.x = evt.pageX - $(this).offset().left; point.y = evt.pageY - $(this).offset().top; test(); }); $("canvas").dblclick(function(evt) { triangle = randomTriangle(); test(); }); test(); function test() { var result = ptInTriangle(point, triangle.a, triangle.b, triangle.c); var info = "point = (" + point.x + "," + point.y + ")\n"; info += "triangle.a = (" + triangle.a.x + "," + triangle.a.y + ")\n"; info += "triangle.b = (" + triangle.b.x + "," + triangle.b.y + ")\n"; info += "triangle.c = (" + triangle.c.x + "," + triangle.c.y + ")\n"; info += "result = " + (result ? "true" : "false"); $("#result").text(info); render(); } function ptInTriangle(p, p0, p1, p2) { var A = 1/2 * (-p1.y * p2.x + p0.y * (-p1.x + p2.x) + p0.x * (p1.y - p2.y) + p1.x * p2.y); var sign = A < 0 ? -1 : 1; var s = (p0.y * p2.x - p0.x * p2.y + (p2.y - p0.y) * p.x + (p0.x - p2.x) * p.y) * sign; var t = (p0.x * p1.y - p0.y * p1.x + (p0.y - p1.y) * p.x + (p1.x - p0.x) * p.y) * sign; return s > 0 && t > 0 && (s + t) < 2 * A * sign; } function render() { ctx.fillStyle = "#CCC"; ctx.fillRect(0, 0, 500, 500); drawTriangle(triangle.a, triangle.b, triangle.c); drawPoint(point); } function drawTriangle(p0, p1, p2) { ctx.fillStyle = "#999"; ctx.beginPath(); ctx.moveTo(p0.x, p0.y); ctx.lineTo(p1.x, p1.y); ctx.lineTo(p2.x, p2.y); ctx.closePath(); ctx.fill(); ctx.fillStyle = "#000"; ctx.font = "12px monospace"; ctx.fillText("1", p0.x, p0.y); ctx.fillText("2", p1.x, p1.y); ctx.fillText("3", p2.x, p2.y); } function drawPoint(p) { ctx.fillStyle = "#F00"; ctx.beginPath(); ctx.arc(p.x, p.y, 5, 0, 2 * Math.PI); ctx.fill(); } function rand(min, max) { return Math.floor(Math.random() * (max - min + 1)) + min; } function randomTriangle() { return { a: { x: rand(0, W), y: rand(0, H) }, b: { x: rand(0, W), y: rand(0, H) }, c: { x: rand(0, W), y: rand(0, H) } }; } <script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script> <pre>Click: place the point. Double click: random triangle.</pre> <pre id="result"></pre> <canvas width="500" height="500"></canvas>

导致:

变量“召回”:30 可变存储:7 补充:4 减法:8 乘法:6 部门:没有 比较:4

这与Kornel Kisielewicz解决方案(25次召回，1次存储，15次减法，6次乘法，5次比较)相比非常好，如果需要顺时针/逆时针检测(它本身需要6次召回，1次加法，2次减法，2次乘法和1次比较，使用解析解行列式，如rhgb所指出的)，可能会更好。

我在JavaScript中改编的高性能代码(文章如下):

function pointInTriangle (p, p0, p1, p2) {
  return (((p1.y - p0.y) * (p.x - p0.x) - (p1.x - p0.x) * (p.y - p0.y)) | ((p2.y - p1.y) * (p.x - p1.x) - (p2.x - p1.x) * (p.y - p1.y)) | ((p0.y - p2.y) * (p.x - p2.x) - (p0.x - p2.x) * (p.y - p2.y))) >= 0;
}

pointInTriangle(p, p0, p1, p2) -用于逆时针方向的三角形 pointInTriangle(p, p0, p1, p2) -用于顺时针三角形

在jsFiddle(包括性能测试)中，在一个单独的函数中也有缠绕检查。或按下面的“运行代码片段”

var ctx = $("canvas")[0].getContext("2d"); var W = 500; var H = 500; var point = { x: W / 2, y: H / 2 }; var triangle = randomTriangle(); $("canvas").click(function(evt) { point.x = evt.pageX - $(this).offset().left; point.y = evt.pageY - $(this).offset().top; test(); }); $("canvas").dblclick(function(evt) { triangle = randomTriangle(); test(); }); document.querySelector('#performance').addEventListener('click', _testPerformance); test(); function test() { var result = checkClockwise(triangle.a, triangle.b, triangle.c) ? pointInTriangle(point, triangle.a, triangle.c, triangle.b) : pointInTriangle(point, triangle.a, triangle.b, triangle.c); var info = "point = (" + point.x + "," + point.y + ")\n"; info += "triangle.a = (" + triangle.a.x + "," + triangle.a.y + ")\n"; info += "triangle.b = (" + triangle.b.x + "," + triangle.b.y + ")\n"; info += "triangle.c = (" + triangle.c.x + "," + triangle.c.y + ")\n"; info += "result = " + (result ? "true" : "false"); $("#result").text(info); render(); } function _testPerformance () { var px = [], py = [], p0x = [], p0y = [], p1x = [], p1y = [], p2x = [], p2y = [], p = [], p0 = [], p1 = [], p2 = []; for(var i = 0; i < 1000000; i++) { p[i] = {x: Math.random() * 100, y: Math.random() * 100}; p0[i] = {x: Math.random() * 100, y: Math.random() * 100}; p1[i] = {x: Math.random() * 100, y: Math.random() * 100}; p2[i] = {x: Math.random() * 100, y: Math.random() * 100}; } console.time('optimal: pointInTriangle'); for(var i = 0; i < 1000000; i++) { pointInTriangle(p[i], p0[i], p1[i], p2[i]); } console.timeEnd('optimal: pointInTriangle'); console.time('original: ptInTriangle'); for(var i = 0; i < 1000000; i++) { ptInTriangle(p[i], p0[i], p1[i], p2[i]); } console.timeEnd('original: ptInTriangle'); } function pointInTriangle (p, p0, p1, p2) { return (((p1.y - p0.y) * (p.x - p0.x) - (p1.x - p0.x) * (p.y - p0.y)) | ((p2.y - p1.y) * (p.x - p1.x) - (p2.x - p1.x) * (p.y - p1.y)) | ((p0.y - p2.y) * (p.x - p2.x) - (p0.x - p2.x) * (p.y - p2.y))) >= 0; } function ptInTriangle(p, p0, p1, p2) { var s = (p0.y * p2.x - p0.x * p2.y + (p2.y - p0.y) * p.x + (p0.x - p2.x) * p.y); var t = (p0.x * p1.y - p0.y * p1.x + (p0.y - p1.y) * p.x + (p1.x - p0.x) * p.y); if (s <= 0 || t <= 0) return false; var A = (-p1.y * p2.x + p0.y * (-p1.x + p2.x) + p0.x * (p1.y - p2.y) + p1.x * p2.y); return (s + t) < A; } function render() { ctx.fillStyle = "#CCC"; ctx.fillRect(0, 0, 500, 500); drawTriangle(triangle.a, triangle.b, triangle.c); drawPoint(point); } function checkClockwise(p0, p1, p2) { var A = (-p1.y * p2.x + p0.y * (-p1.x + p2.x) + p0.x * (p1.y - p2.y) + p1.x * p2.y); return A > 0; } function drawTriangle(p0, p1, p2) { ctx.fillStyle = "#999"; ctx.beginPath(); ctx.moveTo(p0.x, p0.y); ctx.lineTo(p1.x, p1.y); ctx.lineTo(p2.x, p2.y); ctx.closePath(); ctx.fill(); ctx.fillStyle = "#000"; ctx.font = "12px monospace"; ctx.fillText("1", p0.x, p0.y); ctx.fillText("2", p1.x, p1.y); ctx.fillText("3", p2.x, p2.y); } function drawPoint(p) { ctx.fillStyle = "#F00"; ctx.beginPath(); ctx.arc(p.x, p.y, 5, 0, 2 * Math.PI); ctx.fill(); } function rand(min, max) { return Math.floor(Math.random() * (max - min + 1)) + min; } function randomTriangle() { return { a: { x: rand(0, W), y: rand(0, H) }, b: { x: rand(0, W), y: rand(0, H) }, c: { x: rand(0, W), y: rand(0, H) } }; } <script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script> <button id="performance">Run performance test (open console)</button> <pre>Click: place the point. Double click: random triangle.</pre> <pre id="result"></pre> <canvas width="500" height="500"></canvas>

受此启发: http://www.phatcode.net/articles.php?id=459

一般来说，最简单(也是最优)的算法是检查由边创建的半平面的哪一边是点。

以下是关于GameDev的一些高质量信息，包括性能问题。

这里有一些代码让你开始:

float sign (fPoint p1, fPoint p2, fPoint p3)
{
    return (p1.x - p3.x) * (p2.y - p3.y) - (p2.x - p3.x) * (p1.y - p3.y);
}

bool PointInTriangle (fPoint pt, fPoint v1, fPoint v2, fPoint v3)
{
    float d1, d2, d3;
    bool has_neg, has_pos;

    d1 = sign(pt, v1, v2);
    d2 = sign(pt, v2, v3);
    d3 = sign(pt, v3, v1);

    has_neg = (d1 < 0) || (d2 < 0) || (d3 < 0);
    has_pos = (d1 > 0) || (d2 > 0) || (d3 > 0);

    return !(has_neg && has_pos);
}

If you know the co-ordinates of the three vertices and the co-ordinates of the specific point, then you can get the area of the complete triangle. Afterwards, calculate the area of the three triangle segments (one point being the point given and the other two being any two vertices of the triangle). Thus, you will get the area of the three triangle segments. If the sum of these areas are equal to the total area (that you got previously), then, the point should be inside the triangle. Otherwise, the point is not inside the triangle. This should work. If there are any issues, let me know. Thank you.

一个简单的方法是:

找出连接 分别指向三角形的三个点 顶点和夹角之和 这些向量。如果它们的和 角度是2*那么点是 在三角形里面。

两个解释替代方案的好网站是:

黑卒和沃尔夫勒姆