请看图片。如何获得两条线的交点(即绿色圆角点)?我想裁剪图像的内部部分。结束路径是该行中的任何位置。
context = UIGraphicsGetCurrentContext();
CGContextBeginPath(context);
CGContextSetLineWidth(context, 1.0 * self.scale);
CGContextSetLineCap(context, kCGLineCapRound);
[[UIColor redColor] setStroke];
CGPoint firstPoint = CGPointFromString([self.touchPoints objectAtIndex:0]);
CGContextMoveToPoint(context, firstPoint.x, firstPoint.y);
for (NSString *pointString in self.touchPoints) {
CGPoint point = CGPointFromString(pointString);
CGContextAddLineToPoint(context, point.x, point.y);
}
CGContextStrokePath(context);
此代码用于绘制线条。线条画工作正常,裁剪也工作正常......但交点是我的主要问题。请帮我。
最佳答案
想法,检查以firSTLine<>laSTLine、firSTLine<>secondlaSTLine ...firSTLine<>thirdline => secondaryline<>laSTLine等开头的交集。这应该会给你最外面的交集。
以下代码未经测试,但应该可以帮助您解决问题。
typedef struct {
CGPoint startPoint;
CGPoint endPoint;
} Line;
#define CGPointNULL CGPointMake(NAN, NAN)
#define Line(_i_) {CGPointFromString(touchPoints[_i_-1]), CGPointFromString(touchPoints[_i_])};
CGPoint LineIntersects(Line *first, Line *second) {
int x1 = first->startPoint.x; int y1 = first->startPoint.y;
int x2 = first->endPoint.x; int y2 = first->endPoint.y;
int x3 = second->startPoint.x; int y3 = second->startPoint.y;
int x4 = second->endPoint.x; int y4 = second->endPoint.y;
int d = (x1-x2)*(y3-y4) - (y1-y2)*(x3-x4);
if (d == 0) return CGPointNULL;
int xi = ((x3-x4)*(x1*y2-y1*x2)-(x1-x2)*(x3*y4-y3*x4))/d;
int yi = ((y3-y4)*(x1*y2-y1*x2)-(y1-y2)*(x3*y4-y3*x4))/d;
return CGPointMake(xi,yi);
}
static inline BOOL CGPointIsValid(CGPoint p) {
return (p.x != NAN && p.y != NAN);
}
- (CGPoint)mostOuterIntersection:(NSArray *)touchPoints {
CGPoint intersection = CGPointNULL;
int touchCount = [touchPoints count];
for(int i = 1; i<touchCount; i++) {
Line first = Line(i);
for(int j = touchCount-1; j>i+1; j--) {
Line last = Line(j);
intersection = LineIntersects(&first, &last);
if(CGPointIsValid(intersection)) {
break;
}
}
}
return intersection;
}
关于iphone - 如何在iPhone SDK中找到闭合路径(两条线交点)?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/12909008/