让说，我们有两个矩形，用自己的左下和右上边角定义。例如： Rect1的（X1，Y1）（X2，Y2）和 RECT2（X3，Y3）（X4，Y4）。 我试图找到相交矩形的坐标（左下和右上）。

Let say that we have two rectangles, defined with their bottom-left and top-right corners. For example: rect1 (x1, y1)(x2, y2) and rect2 (x3, y3)(x4, y4). I'm trying to find the coordinates(bottom-left and top-right) of the intersected rectangle.

任何想法，算法，伪code，将大大AP preciated。

Any ideas, algorithm, pseudo code, would be greatly appreciated.

P.S。我发现了类似的问题，但他们只检查是否2矩形相交。

p.s. I found similar questions but they check only if 2 rectangle intersect.

推荐答案

如果输入的矩形进行归一化，即你已经知道， X1＆LT; X2 ， Y1＆LT; Y2 （与同为第二个矩形），那么所有你需要做的就是计算

If the input rectangles are normalized, i.e. you already know that x1 < x2, y1 < y2 (and the same for the second rectangle), then all you need to do is calculate

int x5 = max(x1, x3);
int y5 = max(y1, y3);
int x6 = min(x2, x4);
int y6 = min(y2, y4);

和它会给你的交集为矩形（X5，Y5） - （5233，Y6）。如果原来的矩形不相交，其结果将是一个堕落的矩形（用 X5＆GT; = 5233 和/或 Y5＆GT; = Y6 ），您可以轻松地检查。

and it will give you your intersection as rectangle (x5, y5)-(x6, y6). If the original rectangles do not intersect, the result will be a "degenerate" rectangle (with x5 >= x6 and/or y5 >= y6), which you can easily check for.

P.S。像往常一样，小的细节将取决于你是否有考虑的接触的矩形，如图相交。

P.S. As usual, small details will depend on whether you have to consider touching rectangles as intersecting.