我如何确定两条直线是否相交，如果相交，在x,y点处?
当前回答
人们似乎对Gavin的答案很感兴趣，cortijon在评论中提出了一个javascript版本，iMalc提供了一个计算量略少的版本。一些人指出了各种代码建议的缺点，另一些人则评论了一些代码建议的效率。
iMalc通过Gavin的答案提供的算法是我目前在一个javascript项目中使用的算法，我只是想在这里提供一个清理过的版本，如果它可以帮助到任何人的话。
// Some variables for reuse, others may do this differently
var p0x, p1x, p2x, p3x, ix,
p0y, p1y, p2y, p3y, iy,
collisionDetected;
// do stuff, call other functions, set endpoints...
// note: for my purpose I use |t| < |d| as opposed to
// |t| <= |d| which is equivalent to 0 <= t < 1 rather than
// 0 <= t <= 1 as in Gavin's answer - results may vary
var lineSegmentIntersection = function(){
var d, dx1, dx2, dx3, dy1, dy2, dy3, s, t;
dx1 = p1x - p0x; dy1 = p1y - p0y;
dx2 = p3x - p2x; dy2 = p3y - p2y;
dx3 = p0x - p2x; dy3 = p0y - p2y;
collisionDetected = 0;
d = dx1 * dy2 - dx2 * dy1;
if(d !== 0){
s = dx1 * dy3 - dx3 * dy1;
if((s <= 0 && d < 0 && s >= d) || (s >= 0 && d > 0 && s <= d)){
t = dx2 * dy3 - dx3 * dy2;
if((t <= 0 && d < 0 && t > d) || (t >= 0 && d > 0 && t < d)){
t = t / d;
collisionDetected = 1;
ix = p0x + t * dx1;
iy = p0y + t * dy1;
}
}
}
};
其他回答
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。
上面有很多解决方案，但我认为下面的解决方案很简单，很容易理解。
矢量AB和矢量CD相交当且仅当
端点a和b在线段CD的两边。 端点c和d在线段AB的对边。
更具体地说，a和b在线段CD的对面当且仅当两个三元组中有一个是逆时针顺序的。
Intersect(a, b, c, d)
if CCW(a, c, d) == CCW(b, c, d)
return false;
else if CCW(a, b, c) == CCW(a, b, d)
return false;
else
return true;
这里的CCW代表逆时针，根据点的方向返回真/假。
来源:http://compgeom.cs.uiuc.edu/~jeffe/teaching/373/notes/x06-sweepline.pdf 第二页
有一个很好的方法来解决这个问题就是用向量叉乘。定义二维向量叉乘v × w为vx wy−vy wx。
假设这两条线段从p到p + r，从q到q + s。那么第一行上的任意点都可以表示为p + t r(对于标量参数t)，第二行上的任意点可以表示为q + u s(对于标量参数u)。
如果t和u满足以下条件，两条直线相交:
P + t r = q + u s
两边叉乘s，得到
(p + r) × s = (q + u s) × s
由于s × s = 0，这意味着
T (r × s) = (q−p) × s
因此，求解t:
T = (q−p) × s / (r × s)
同样地，我们可以解出u:
(p + r) × r = (q + u s) × r U (s × r) = (p−q) × r U = (p−q) × r / (s × r)
为了减少计算步骤，可以方便地将其重写为以下形式(记住s × r =−r × s):
U = q−p × r / (r × s)
现在有四种情况:
If r × s = 0 and (q − p) × r = 0, then the two lines are collinear. In this case, express the endpoints of the second segment (q and q + s) in terms of the equation of the first line segment (p + t r): t0 = (q − p) · r / (r · r) t1 = (q + s − p) · r / (r · r) = t0 + s · r / (r · r) If the interval between t0 and t1 intersects the interval [0, 1] then the line segments are collinear and overlapping; otherwise they are collinear and disjoint. Note that if s and r point in opposite directions, then s · r < 0 and so the interval to be checked is [t1, t0] rather than [t0, t1]. If r × s = 0 and (q − p) × r ≠ 0, then the two lines are parallel and non-intersecting. If r × s ≠ 0 and 0 ≤ t ≤ 1 and 0 ≤ u ≤ 1, the two line segments meet at the point p + t r = q + u s. Otherwise, the two line segments are not parallel but do not intersect.
来源:该方法是3D线相交算法的2维专门化，来自Ronald Goldman发表在Graphics Gems，第304页的文章“三条线在三维空间中的相交”。在三维空间中，通常的情况是直线是倾斜的(既不平行也不相交)，在这种情况下，该方法给出了两条直线最接近的点。
我试过其中一些答案，但它们对我不起作用(对不起伙计们);在网上搜索之后，我找到了这个。
对他的代码做了一点修改，我现在有了这个函数，它将返回交点，如果没有找到交点，它将返回- 1,1。
Public Function intercetion(ByVal ax As Integer, ByVal ay As Integer, ByVal bx As Integer, ByVal by As Integer, ByVal cx As Integer, ByVal cy As Integer, ByVal dx As Integer, ByVal dy As Integer) As Point
'// Determines the intersection point of the line segment defined by points A and B
'// with the line segment defined by points C and D.
'//
'// Returns YES if the intersection point was found, and stores that point in X,Y.
'// Returns NO if there is no determinable intersection point, in which case X,Y will
'// be unmodified.
Dim distAB, theCos, theSin, newX, ABpos As Double
'// Fail if either line segment is zero-length.
If ax = bx And ay = by Or cx = dx And cy = dy Then Return New Point(-1, -1)
'// Fail if the segments share an end-point.
If ax = cx And ay = cy Or bx = cx And by = cy Or ax = dx And ay = dy Or bx = dx And by = dy Then Return New Point(-1, -1)
'// (1) Translate the system so that point A is on the origin.
bx -= ax
by -= ay
cx -= ax
cy -= ay
dx -= ax
dy -= ay
'// Discover the length of segment A-B.
distAB = Math.Sqrt(bx * bx + by * by)
'// (2) Rotate the system so that point B is on the positive X axis.
theCos = bx / distAB
theSin = by / distAB
newX = cx * theCos + cy * theSin
cy = cy * theCos - cx * theSin
cx = newX
newX = dx * theCos + dy * theSin
dy = dy * theCos - dx * theSin
dx = newX
'// Fail if segment C-D doesn't cross line A-B.
If cy < 0 And dy < 0 Or cy >= 0 And dy >= 0 Then Return New Point(-1, -1)
'// (3) Discover the position of the intersection point along line A-B.
ABpos = dx + (cx - dx) * dy / (dy - cy)
'// Fail if segment C-D crosses line A-B outside of segment A-B.
If ABpos < 0 Or ABpos > distAB Then Return New Point(-1, -1)
'// (4) Apply the discovered position to line A-B in the original coordinate system.
'*X=Ax+ABpos*theCos
'*Y=Ay+ABpos*theSin
'// Success.
Return New Point(ax + ABpos * theCos, ay + ABpos * theSin)
End Function
人们似乎对Gavin的答案很感兴趣，cortijon在评论中提出了一个javascript版本，iMalc提供了一个计算量略少的版本。一些人指出了各种代码建议的缺点，另一些人则评论了一些代码建议的效率。
iMalc通过Gavin的答案提供的算法是我目前在一个javascript项目中使用的算法，我只是想在这里提供一个清理过的版本，如果它可以帮助到任何人的话。
// Some variables for reuse, others may do this differently
var p0x, p1x, p2x, p3x, ix,
p0y, p1y, p2y, p3y, iy,
collisionDetected;
// do stuff, call other functions, set endpoints...
// note: for my purpose I use |t| < |d| as opposed to
// |t| <= |d| which is equivalent to 0 <= t < 1 rather than
// 0 <= t <= 1 as in Gavin's answer - results may vary
var lineSegmentIntersection = function(){
var d, dx1, dx2, dx3, dy1, dy2, dy3, s, t;
dx1 = p1x - p0x; dy1 = p1y - p0y;
dx2 = p3x - p2x; dy2 = p3y - p2y;
dx3 = p0x - p2x; dy3 = p0y - p2y;
collisionDetected = 0;
d = dx1 * dy2 - dx2 * dy1;
if(d !== 0){
s = dx1 * dy3 - dx3 * dy1;
if((s <= 0 && d < 0 && s >= d) || (s >= 0 && d > 0 && s <= d)){
t = dx2 * dy3 - dx3 * dy2;
if((t <= 0 && d < 0 && t > d) || (t >= 0 && d > 0 && t < d)){
t = t / d;
collisionDetected = 1;
ix = p0x + t * dx1;
iy = p0y + t * dy1;
}
}
}
};
