C和Objective-C

基于Gareth Rees的回答

const AGKLine AGKLineZero = (AGKLine){(CGPoint){0.0, 0.0}, (CGPoint){0.0, 0.0}};

AGKLine AGKLineMake(CGPoint start, CGPoint end)
{
    return (AGKLine){start, end};
}

double AGKLineLength(AGKLine l)
{
    return CGPointLengthBetween_AGK(l.start, l.end);
}

BOOL AGKLineIntersection(AGKLine l1, AGKLine l2, CGPoint *out_pointOfIntersection)
{
    // http://stackoverflow.com/a/565282/202451

    CGPoint p = l1.start;
    CGPoint q = l2.start;
    CGPoint r = CGPointSubtract_AGK(l1.end, l1.start);
    CGPoint s = CGPointSubtract_AGK(l2.end, l2.start);
    
    double s_r_crossProduct = CGPointCrossProductZComponent_AGK(r, s);
    double t = CGPointCrossProductZComponent_AGK(CGPointSubtract_AGK(q, p), s) / s_r_crossProduct;
    double u = CGPointCrossProductZComponent_AGK(CGPointSubtract_AGK(q, p), r) / s_r_crossProduct;
    
    if(t < 0 || t > 1.0 || u < 0 || u > 1.0)
    {
        if(out_pointOfIntersection != NULL)
        {
            *out_pointOfIntersection = CGPointZero;
        }
        return NO;
    }
    else
    {
        if(out_pointOfIntersection != NULL)
        {
            CGPoint i = CGPointAdd_AGK(p, CGPointMultiply_AGK(r, t));
            *out_pointOfIntersection = i;
        }
        return YES;
    }
}

CGFloat CGPointCrossProductZComponent_AGK(CGPoint v1, CGPoint v2)
{
    return v1.x * v2.y - v1.y * v2.x;
}

CGPoint CGPointSubtract_AGK(CGPoint p1, CGPoint p2)
{
    return (CGPoint){p1.x - p2.x, p1.y - p2.y};
}

CGPoint CGPointAdd_AGK(CGPoint p1, CGPoint p2)
{
    return (CGPoint){p1.x + p2.x, p1.y + p2.y};
}

CGFloat CGPointCrossProductZComponent_AGK(CGPoint v1, CGPoint v2)
{
    return v1.x * v2.y - v1.y * v2.x;
}

CGPoint CGPointMultiply_AGK(CGPoint p1, CGFloat factor)
{
    return (CGPoint){p1.x * factor, p1.y * factor};
}

许多函数和结构都是私有的，但是你应该很容易就能知道发生了什么。 这是公开的在这个回购https://github.com/hfossli/AGGeometryKit/

以下是对加文回答的改进。马普的解决方案也类似，但都没有推迟分割。

这实际上也是Gareth Rees的答案的一个实际应用，因为向量积在2D中的等价是补点积，这段代码用了其中的三个。切换到3D并使用叉积，在最后插入s和t，结果是3D中直线之间的两个最近点。 不管怎样，2D解:

int get_line_intersection(float p0_x, float p0_y, float p1_x, float p1_y, 
    float p2_x, float p2_y, float p3_x, float p3_y, float *i_x, float *i_y)
{
    float s02_x, s02_y, s10_x, s10_y, s32_x, s32_y, s_numer, t_numer, denom, t;
    s10_x = p1_x - p0_x;
    s10_y = p1_y - p0_y;
    s32_x = p3_x - p2_x;
    s32_y = p3_y - p2_y;

    denom = s10_x * s32_y - s32_x * s10_y;
    if (denom == 0)
        return 0; // Collinear
    bool denomPositive = denom > 0;

    s02_x = p0_x - p2_x;
    s02_y = p0_y - p2_y;
    s_numer = s10_x * s02_y - s10_y * s02_x;
    if ((s_numer < 0) == denomPositive)
        return 0; // No collision

    t_numer = s32_x * s02_y - s32_y * s02_x;
    if ((t_numer < 0) == denomPositive)
        return 0; // No collision

    if (((s_numer > denom) == denomPositive) || ((t_numer > denom) == denomPositive))
        return 0; // No collision
    // Collision detected
    t = t_numer / denom;
    if (i_x != NULL)
        *i_x = p0_x + (t * s10_x);
    if (i_y != NULL)
        *i_y = p0_y + (t * s10_y);

    return 1;
}

基本上，它将除法延迟到最后一刻，并将大多数测试移动到某些计算完成之前，从而增加了早期退出。最后，它还避免了直线平行时的除零情况。

您可能还想考虑使用ε检验，而不是与零比较。非常接近平行的线会产生稍微偏离的结果。这不是一个bug，这是浮点数学的一个限制。

C和Objective-C

基于Gareth Rees的回答

const AGKLine AGKLineZero = (AGKLine){(CGPoint){0.0, 0.0}, (CGPoint){0.0, 0.0}};

AGKLine AGKLineMake(CGPoint start, CGPoint end)
{
    return (AGKLine){start, end};
}

double AGKLineLength(AGKLine l)
{
    return CGPointLengthBetween_AGK(l.start, l.end);
}

BOOL AGKLineIntersection(AGKLine l1, AGKLine l2, CGPoint *out_pointOfIntersection)
{
    // http://stackoverflow.com/a/565282/202451

    CGPoint p = l1.start;
    CGPoint q = l2.start;
    CGPoint r = CGPointSubtract_AGK(l1.end, l1.start);
    CGPoint s = CGPointSubtract_AGK(l2.end, l2.start);
    
    double s_r_crossProduct = CGPointCrossProductZComponent_AGK(r, s);
    double t = CGPointCrossProductZComponent_AGK(CGPointSubtract_AGK(q, p), s) / s_r_crossProduct;
    double u = CGPointCrossProductZComponent_AGK(CGPointSubtract_AGK(q, p), r) / s_r_crossProduct;
    
    if(t < 0 || t > 1.0 || u < 0 || u > 1.0)
    {
        if(out_pointOfIntersection != NULL)
        {
            *out_pointOfIntersection = CGPointZero;
        }
        return NO;
    }
    else
    {
        if(out_pointOfIntersection != NULL)
        {
            CGPoint i = CGPointAdd_AGK(p, CGPointMultiply_AGK(r, t));
            *out_pointOfIntersection = i;
        }
        return YES;
    }
}

CGFloat CGPointCrossProductZComponent_AGK(CGPoint v1, CGPoint v2)
{
    return v1.x * v2.y - v1.y * v2.x;
}

CGPoint CGPointSubtract_AGK(CGPoint p1, CGPoint p2)
{
    return (CGPoint){p1.x - p2.x, p1.y - p2.y};
}

CGPoint CGPointAdd_AGK(CGPoint p1, CGPoint p2)
{
    return (CGPoint){p1.x + p2.x, p1.y + p2.y};
}

CGFloat CGPointCrossProductZComponent_AGK(CGPoint v1, CGPoint v2)
{
    return v1.x * v2.y - v1.y * v2.x;
}

CGPoint CGPointMultiply_AGK(CGPoint p1, CGFloat factor)
{
    return (CGPoint){p1.x * factor, p1.y * factor};
}

许多函数和结构都是私有的，但是你应该很容易就能知道发生了什么。 这是公开的在这个回购https://github.com/hfossli/AGGeometryKit/

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;
}

这是基于Gareth Ree的回答。它还返回线段重叠的情况。用c++编写的V是一个简单的向量类。其中二维中两个向量的外积返回一个标量。通过了学校自动测试系统的测试。

//Required input point must be colinear with the line
bool on_segment(const V& p, const LineSegment& l)
{
    //If a point is on the line, the sum of the vectors formed by the point to the line endpoints must be equal
    V va = p - l.pa;
    V vb = p - l.pb;
    R ma = va.magnitude();
    R mb = vb.magnitude();
    R ml = (l.pb - l.pa).magnitude();
    R s = ma + mb;
    bool r = s <= ml + epsilon;
    return r;
}

//Compute using vector math
// Returns 0 points if the lines do not intersect or overlap
// Returns 1 point if the lines intersect
//  Returns 2 points if the lines overlap, contain the points where overlapping start starts and stop
std::vector<V> intersect(const LineSegment& la, const LineSegment& lb)
{
    std::vector<V> r;

    //http://stackoverflow.com/questions/563198/how-do-you-detect-where-two-line-segments-intersect
    V oa, ob, da, db; //Origin and direction vectors
    R sa, sb; //Scalar values
    oa = la.pa;
    da = la.pb - la.pa;
    ob = lb.pa;
    db = lb.pb - lb.pa;

    if (da.cross(db) == 0 && (ob - oa).cross(da) == 0) //If colinear
    {
        if (on_segment(lb.pa, la) && on_segment(lb.pb, la))
        {
            r.push_back(lb.pa);
            r.push_back(lb.pb);
            dprintf("colinear, overlapping\n");
            return r;
        }

        if (on_segment(la.pa, lb) && on_segment(la.pb, lb))
        {
            r.push_back(la.pa);
            r.push_back(la.pb);
            dprintf("colinear, overlapping\n");
            return r;
        }

        if (on_segment(la.pa, lb))
            r.push_back(la.pa);

        if (on_segment(la.pb, lb))
            r.push_back(la.pb);

        if (on_segment(lb.pa, la))
            r.push_back(lb.pa);

        if (on_segment(lb.pb, la))
            r.push_back(lb.pb);

        if (r.size() == 0)
            dprintf("colinear, non-overlapping\n");
        else
            dprintf("colinear, overlapping\n");

        return r;
    }

    if (da.cross(db) == 0 && (ob - oa).cross(da) != 0)
    {
        dprintf("parallel non-intersecting\n");
        return r;
    }

    //Math trick db cross db == 0, which is a single scalar in 2D.
    //Crossing both sides with vector db gives:
    sa = (ob - oa).cross(db) / da.cross(db);

    //Crossing both sides with vector da gives
    sb = (oa - ob).cross(da) / db.cross(da);

    if (0 <= sa && sa <= 1 && 0 <= sb && sb <= 1)
    {
        dprintf("intersecting\n");
        r.push_back(oa + da * sa);
        return r;
    }

    dprintf("non-intersecting, non-parallel, non-colinear, non-overlapping\n");
    return r;
}

基于@Gareth Rees的回答，Python版本:

import numpy as np

def np_perp( a ) :
    b = np.empty_like(a)
    b[0] = a[1]
    b[1] = -a[0]
    return b

def np_cross_product(a, b):
    return np.dot(a, np_perp(b))

def np_seg_intersect(a, b, considerCollinearOverlapAsIntersect = False):
    # https://stackoverflow.com/questions/563198/how-do-you-detect-where-two-line-segments-intersect/565282#565282
    # http://www.codeproject.com/Tips/862988/Find-the-intersection-point-of-two-line-segments
    r = a[1] - a[0]
    s = b[1] - b[0]
    v = b[0] - a[0]
    num = np_cross_product(v, r)
    denom = np_cross_product(r, s)
    # If r x s = 0 and (q - p) x r = 0, then the two lines are collinear.
    if np.isclose(denom, 0) and np.isclose(num, 0):
        # 1. If either  0 <= (q - p) * r <= r * r or 0 <= (p - q) * s <= * s
        # then the two lines are overlapping,
        if(considerCollinearOverlapAsIntersect):
            vDotR = np.dot(v, r)
            aDotS = np.dot(-v, s)
            if (0 <= vDotR  and vDotR <= np.dot(r,r)) or (0 <= aDotS  and aDotS <= np.dot(s,s)):
                return True
        # 2. If neither 0 <= (q - p) * r = r * r nor 0 <= (p - q) * s <= s * s
        # then the two lines are collinear but disjoint.
        # No need to implement this expression, as it follows from the expression above.
        return None
    if np.isclose(denom, 0) and not np.isclose(num, 0):
        # Parallel and non intersecting
        return None
    u = num / denom
    t = np_cross_product(v, s) / denom
    if u >= 0 and u <= 1 and t >= 0 and t <= 1:
        res = b[0] + (s*u)
        return res
    # Otherwise, the two line segments are not parallel but do not intersect.
    return None