更新时间:2023-11-26 18:12:10
以下算法可用于检查两个(旋转或以其他方式转换的)视图是否重叠:
The following algorithm can be used to check if two (rotated or otherwise transformed) views overlap:
[view convertPoint:point toView:nil]
转换两个视图的4个边界点
到通用坐标系(窗口坐标).[view convertPoint:point toView:nil]
to convert the 4 boundary points of both views
to a common coordinate system (the window coordinates).此: http://www.geometrictools.com/Documentation/MethodOfSeparatingAxes.pdf 是对包含伪代码的算法,可以通过谷歌搜索分离轴定理"找到更多的方法.
This: http://www.geometrictools.com/Documentation/MethodOfSeparatingAxes.pdf is another description of the algorithm containing pseudo-code, more can be found by googling for "Separating Axis Theorem".
更新:我试图为分离轴定理"创建一个Objective-C方法,这就是我所得到的.到现在为止,我只进行了几次测试,所以希望不会出现太多错误.
Update: I have tried to create a Objective-C method for the "Separating Axis Theorem", and this is what I got. Up to now, I did only a few tests, so I hope that there are not too many errors.
- (BOOL)convexPolygon:(CGPoint *)poly1 count:(int)count1 intersectsWith:(CGPoint *)poly2 count:(int)count2;
测试2个凸多边形是否相交.这两个多边形均以顶点的CGPoint
数组形式给出.
tests if 2 convex polygons intersect. Both polygons are given as a CGPoint
array of the vertices.
- (BOOL)view:(UIView *)view1 intersectsWith:(UIView *)view2
测试(如上所述)两个任意视图是否相交.
tests (as described above) if two arbitrary views intersect.
实施:
- (void)projectionOfPolygon:(CGPoint *)poly count:(int)count onto:(CGPoint)perp min:(CGFloat *)minp max:(CGFloat *)maxp
{
CGFloat minproj = MAXFLOAT;
CGFloat maxproj = -MAXFLOAT;
for (int j = 0; j < count; j++) {
CGFloat proj = poly[j].x * perp.x + poly[j].y * perp.y;
if (proj > maxproj)
maxproj = proj;
if (proj < minproj)
minproj = proj;
}
*minp = minproj;
*maxp = maxproj;
}
-(BOOL)convexPolygon:(CGPoint *)poly1 count:(int)count1 intersectsWith:(CGPoint *)poly2 count:(int)count2
{
for (int i = 0; i < count1; i++) {
// Perpendicular vector for one edge of poly1:
CGPoint p1 = poly1[i];
CGPoint p2 = poly1[(i+1) % count1];
CGPoint perp = CGPointMake(- (p2.y - p1.y), p2.x - p1.x);
// Projection intervals of poly1, poly2 onto perpendicular vector:
CGFloat minp1, maxp1, minp2, maxp2;
[self projectionOfPolygon:poly1 count:count1 onto:perp min:&minp1 max:&maxp1];
[self projectionOfPolygon:poly2 count:count1 onto:perp min:&minp2 max:&maxp2];
// If projections do not overlap then we have a "separating axis"
// which means that the polygons do not intersect:
if (maxp1 < minp2 || maxp2 < minp1)
return NO;
}
// And now the other way around with edges from poly2:
for (int i = 0; i < count2; i++) {
CGPoint p1 = poly2[i];
CGPoint p2 = poly2[(i+1) % count2];
CGPoint perp = CGPointMake(- (p2.y - p1.y), p2.x - p1.x);
CGFloat minp1, maxp1, minp2, maxp2;
[self projectionOfPolygon:poly1 count:count1 onto:perp min:&minp1 max:&maxp1];
[self projectionOfPolygon:poly2 count:count1 onto:perp min:&minp2 max:&maxp2];
if (maxp1 < minp2 || maxp2 < minp1)
return NO;
}
// No separating axis found, then the polygons must intersect:
return YES;
}
- (BOOL)view:(UIView *)view1 intersectsWith:(UIView *)view2
{
CGPoint poly1[4];
CGRect bounds1 = view1.bounds;
poly1[0] = [view1 convertPoint:bounds1.origin toView:nil];
poly1[1] = [view1 convertPoint:CGPointMake(bounds1.origin.x + bounds1.size.width, bounds1.origin.y) toView:nil];
poly1[2] = [view1 convertPoint:CGPointMake(bounds1.origin.x + bounds1.size.width, bounds1.origin.y + bounds1.size.height) toView:nil];
poly1[3] = [view1 convertPoint:CGPointMake(bounds1.origin.x, bounds1.origin.y + bounds1.size.height) toView:nil];
CGPoint poly2[4];
CGRect bounds2 = view2.bounds;
poly2[0] = [view2 convertPoint:bounds2.origin toView:nil];
poly2[1] = [view2 convertPoint:CGPointMake(bounds2.origin.x + bounds2.size.width, bounds2.origin.y) toView:nil];
poly2[2] = [view2 convertPoint:CGPointMake(bounds2.origin.x + bounds2.size.width, bounds2.origin.y + bounds2.size.height) toView:nil];
poly2[3] = [view2 convertPoint:CGPointMake(bounds2.origin.x, bounds2.origin.y + bounds2.size.height) toView:nil];
return [self convexPolygon:poly1 count:4 intersectsWith:poly2 count:4];
}