下面的算法可用於檢查兩個(旋轉或以其它方式轉化的)視圖重疊:
這是:http://www.geometrictools.com/Documentation/MethodOfSeparatingAxes.pdf是包含僞代碼的算法的另一種描述,更多可以通過谷歌搜索「分離軸定理」找到。
更新:我試圖創造了「分離軸定理」一個Objective-C的方法,這是我得到了什麼。到目前爲止,我只做了一些測試,所以我希望沒有太多的錯誤。
- (BOOL)convexPolygon:(CGPoint *)poly1 count:(int)count1 intersectsWith:(CGPoint *)poly2 count:(int)count2;
測試如果2個凸多邊形相交。兩個多邊形均以頂點數組形式給出。
- (BOOL)view:(UIView *)view1 intersectsWith:(UIView *)view2
測試(如上所述)如果兩個任意視圖相交。
實現:
- (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];
}
的算法的另一個描述:http://stackoverflow.com/a/115520/1187415。 –
太棒了!非常感謝! – David
@MartinR我問了一個新的問題[這裏](http://stackoverflow.com/questions/26821725/determine-if-crop-rect-is-entirely-contained-within-rotated-uiview)基於這個問題/回答。想知道你是否可以看一看? – brandonscript