找到重叠间隔的整洁算法是什么？
我确信以前肯定有人问过这个问题，但我没有找到它：我只是找到相关但更难的问题。
我有四个点，代表两条线，如下所示：
A C B D
|------*---|-----+----|-*---+---|----------|
0 10 20 30 40
因此在示例中，AB = {7, 21} 和 CD = {16,26}。 （这些线彼此之间可以有任何关系，也可以有任何大小。）我想知道它们是否重叠，如果重叠的话，重叠多少。 （在这个例子中，答案是 5。）我当前的解决方案涉及一堆复杂的 if/then 步骤，我忍不住认为有一个很好的算术解决方案。有没有？
（PS：真的，我正在做边界框相交，但如果我能在一个维度上得到它，显然另一个维度将是相同的。）
I'm sure this must have been asked before, but I'm not finding it: I'm only finding related, but harder questions.
I've got four points, representing two lines like this:
A C B D
|------*---|-----+----|-*---+---|----------|
0 10 20 30 40
So in the example, AB = {7, 21} and CD = {16,26}. (The lines could be in any relation to each other, and any size.) I want to find out whether or not they overlap, and by how much if so. (In the example, the answer would be 5.) My current solution involves a bunch of complicated if/then steps, and I can't help but think there's a nice arithmetical solution. Is there?
(P.S. Really, I'm doing bounding-box intersection, but if I can get it in one dimension, the other will be the same, obviously.)
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试试这个：
如果您可以假设
a <= b和c <= d：您还可以计算
相交，如下所示：Try this:
If you can assume
a <= bandc <= d:You can also calculate
intersectsas follows:线段是两条相对射线（相反方向的两条半无限直线）的交点。有两条线段相交——结果是所有 4 条射线的交点。因此，您可以将代码分解为三个连续的光线相交：左侧光线的左侧与右侧光线的右侧相交。
（这是陈述现已接受的答案的另一种方式。）
A line segment is the intersection of two opposing rays (two half-infinite lines in opposite directions). You have two line segments to intersect -- the result is the intersection of all 4 rays. So you can factor your code as three successive ray-intersections: the leftward of the left-facing rays intersected with the rightward of the right-facing rays.
(This is another way of stating the now-accepted answer.)