我将Kris的答案移植到JavaScript。在尝试了许多不同的答案后，他给出了正确的观点。我以为我要疯了，因为我没有得到我需要的分数。

function getLineLineCollision(p0, p1, p2, p3) {
    var s1, s2;
    s1 = {x: p1.x - p0.x, y: p1.y - p0.y};
    s2 = {x: p3.x - p2.x, y: p3.y - p2.y};

    var s10_x = p1.x - p0.x;
    var s10_y = p1.y - p0.y;
    var s32_x = p3.x - p2.x;
    var s32_y = p3.y - p2.y;

    var denom = s10_x * s32_y - s32_x * s10_y;

    if(denom == 0) {
        return false;
    }

    var denom_positive = denom > 0;

    var s02_x = p0.x - p2.x;
    var s02_y = p0.y - p2.y;

    var s_numer = s10_x * s02_y - s10_y * s02_x;

    if((s_numer < 0) == denom_positive) {
        return false;
    }

    var t_numer = s32_x * s02_y - s32_y * s02_x;

    if((t_numer < 0) == denom_positive) {
        return false;
    }

    if((s_numer > denom) == denom_positive || (t_numer > denom) == denom_positive) {
        return false;
    }

    var t = t_numer / denom;

    var p = {x: p0.x + (t * s10_x), y: p0.y + (t * s10_y)};
    return p;
}

如果矩形的每条边都是一条线段，并且用户绘制的部分也是一条线段，那么您只需检查用户绘制的线段是否与四条边线段相交。这应该是一个相当简单的练习，给定每个段的起点和终点。

只是想提一下，一个很好的解释和明确的解决方案可以在数字食谱系列中找到。我有这本书的第三版，答案在1117页21.4节。另一种不同命名的解决方案可以在玛丽娜·加夫里洛娃(Marina Gavrilova)的论文中找到。在我看来，她的解决办法要简单一些。

我的实现如下:

bool NuGeometry::IsBetween(const double& x0, const double& x, const double& x1){
   return (x >= x0) && (x <= x1);
}

bool NuGeometry::FindIntersection(const double& x0, const double& y0, 
     const double& x1, const double& y1,
     const double& a0, const double& b0, 
     const double& a1, const double& b1, 
     double& xy, double& ab) {
   // four endpoints are x0, y0 & x1,y1 & a0,b0 & a1,b1
   // returned values xy and ab are the fractional distance along xy and ab
   // and are only defined when the result is true

   bool partial = false;
   double denom = (b0 - b1) * (x0 - x1) - (y0 - y1) * (a0 - a1);
   if (denom == 0) {
      xy = -1;
      ab = -1;
   } else {
      xy = (a0 * (y1 - b1) + a1 * (b0 - y1) + x1 * (b1 - b0)) / denom;
      partial = NuGeometry::IsBetween(0, xy, 1);
      if (partial) {
         // no point calculating this unless xy is between 0 & 1
         ab = (y1 * (x0 - a1) + b1 * (x1 - x0) + y0 * (a1 - x1)) / denom; 
      }
   }
   if ( partial && NuGeometry::IsBetween(0, ab, 1)) {
      ab = 1-ab;
      xy = 1-xy;
      return true;
   }  else return false;
}

我认为这个问题有一个更简单的解决方案。今天我想到了另一个想法，看起来效果不错(至少在2D中)。你所要做的就是计算两条直线的交点，然后检查计算的交点是否在两条线段的边界框内。如果是，两条线段相交。就是这样。

编辑:

这就是我如何计算交集(我不知道我在哪里找到了这个代码片段)

Point3D

来自

System.Windows.Media.Media3D

public static Point3D? Intersection(Point3D start1, Point3D end1, Point3D start2, Point3D end2) {

        double a1 = end1.Y - start1.Y;
        double b1 = start1.X - end1.X;
        double c1 = a1 * start1.X + b1 * start1.Y;

        double a2 = end2.Y - start2.Y;
        double b2 = start2.X - end2.X;
        double c2 = a2 * start2.X + b2 * start2.Y;

        double det = a1 * b2 - a2 * b1;
        if (det == 0) { // lines are parallel
            return null;
        }

        double x = (b2 * c1 - b1 * c2) / det;
        double y = (a1 * c2 - a2 * c1) / det;

        return new Point3D(x, y, 0.0);
    }

这是我的BoundingBox类(为了回答的目的而简化):

public class BoundingBox {
    private Point3D min = new Point3D();
    private Point3D max = new Point3D();

    public BoundingBox(Point3D point) {
        min = point;
        max = point;
    }

    public Point3D Min {
        get { return min; }
        set { min = value; }
    }

    public Point3D Max {
        get { return max; }
        set { max = value; }
    }

    public bool Contains(BoundingBox box) {
        bool contains =
            min.X <= box.min.X && max.X >= box.max.X &&
            min.Y <= box.min.Y && max.Y >= box.max.Y &&
            min.Z <= box.min.Z && max.Z >= box.max.Z;
        return contains;
    }

    public bool Contains(Point3D point) {
        return Contains(new BoundingBox(point));
    }

}

曾经在这里被接受的答案是不正确的(它已经被不接受了，所以万岁!)它不能正确地消除所有非交点。简单地说，它可能有效，但也可能失败，特别是在0和1被认为对h有效的情况下。

考虑以下情况:

直线(4,1)-(5,1)和(0,0)-(0,2)

这两条垂线显然不重叠。

= (4,1) B =(5、1) C = (0, 0) D = (0, 2) E = (1) - (4,1) = (1,0) F = (0, 2) - (0, 0) = (0, 2) P = (0, 1) h =((4,1) -(0, 0))点(0,1)/((0,2)点(0,1))= 0

根据上面的答案，这两条线段在端点处相遇(值为0和1)。该端点为:

(0, 0) + (0, 2) * 0 = (0, 0)

So, apparently the two line segments meet at (0,0), which is on line CD, but not on line AB. So what is going wrong? The answer is that the values of 0 and 1 are not valid and only sometimes HAPPEN to correctly predict endpoint intersection. When the extension of one line (but not the other) would meet the line segment, the algorithm predicts an intersection of line segments, but this is not correct. I imagine that by testing starting with AB vs CD and then also testing with CD vs AB, this problem would be eliminated. Only if both fall between 0 and 1 inclusively can they be said to intersect.

如果你必须预测端点，我建议使用向量叉乘法。

-Dan

我将Kris的答案移植到JavaScript。在尝试了许多不同的答案后，他给出了正确的观点。我以为我要疯了，因为我没有得到我需要的分数。

function getLineLineCollision(p0, p1, p2, p3) {
    var s1, s2;
    s1 = {x: p1.x - p0.x, y: p1.y - p0.y};
    s2 = {x: p3.x - p2.x, y: p3.y - p2.y};

    var s10_x = p1.x - p0.x;
    var s10_y = p1.y - p0.y;
    var s32_x = p3.x - p2.x;
    var s32_y = p3.y - p2.y;

    var denom = s10_x * s32_y - s32_x * s10_y;

    if(denom == 0) {
        return false;
    }

    var denom_positive = denom > 0;

    var s02_x = p0.x - p2.x;
    var s02_y = p0.y - p2.y;

    var s_numer = s10_x * s02_y - s10_y * s02_x;

    if((s_numer < 0) == denom_positive) {
        return false;
    }

    var t_numer = s32_x * s02_y - s32_y * s02_x;

    if((t_numer < 0) == denom_positive) {
        return false;
    }

    if((s_numer > denom) == denom_positive || (t_numer > denom) == denom_positive) {
        return false;
    }

    var t = t_numer / denom;

    var p = {x: p0.x + (t * s10_x), y: p0.y + (t * s10_y)};
    return p;
}