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我在实现用于绘制贝塞尔曲线的代码时遇到问题，该曲线在 iOS 中运行良好。

我想要这个效果。但有接触点。

但我搞错了。这条线的问题

previousCenterPoint = CenterPointOf(new PointF(points.get(0).x,points.get(0).y), previousPoint);

在 iOS 中，使用 currentPoint 我们可以获得当前点。

请建议我如何获得路径的当前轮廓点....

这是我的代码..

public void makeBezierCurve(ArrayList<PointF> points) {
    if (points.size() > 0) {

        if (points.size() < 3) {

            switch (points.size()) {

                case 1:
                    lineTo(points.get(0).x, points.get(0).y);
                case 2:
                    lineTo(points.get(1).x, points.get(1).y);
                default:
                    break;
            }
        } else {


            PointF previousPoint = new PointF(0, 0);
            PointF previousCenterPoint = new PointF(0, 0);
            PointF centerPoint = new PointF(0, 0);
            double centerPointDistance = 0;
            double obliqueAngle = 0;
            PointF previousControlPoint1 = new PointF(0, 0);
            PointF previousControlPoint2 = new PointF(0, 0);
            PointF controlPoint1 = new PointF(0, 0);
            float contractionFactor = 0.7f;

            for (int i = 0; i < points.size(); i++) {

                PointF pointI = points.get(i);

                if (i > 0) {

                    previousCenterPoint = CenterPointOf(new PointF(points.get(0).x, points.get(0).y), previousPoint);
                    centerPoint = CenterPointOf(previousPoint, pointI);

                    centerPointDistance = DistanceBetween(previousCenterPoint, centerPoint);

                    obliqueAngle = ObliqueAngleOfStraightThrough(centerPoint, previousCenterPoint);

                    previousControlPoint2 = new PointF((float) (previousPoint.x - 0.5 * contractionFactor * centerPointDistance * Math.cos(obliqueAngle)), (float) (previousPoint.y - 0.5 * contractionFactor * centerPointDistance * Math.sin(obliqueAngle)));
                    controlPoint1 = new PointF((float) (previousPoint.x + 0.5 * contractionFactor * centerPointDistance * Math.cos(obliqueAngle)), (float) (previousPoint.y + 0.5 * contractionFactor * centerPointDistance * Math.sin(obliqueAngle)));
                }

                if (i == 1) {
                    quadTo(previousControlPoint2.x, previousControlPoint2.y, previousPoint.x, previousPoint.y);
                } else if (i >= 2 && i < points.size() - 1) {
                    cubicTo(previousControlPoint1.x, previousControlPoint1.y, previousControlPoint2.x, previousControlPoint2.y, previousPoint.x, previousPoint.y);
                } else if (i == points.size() - 1) {
                    cubicTo(previousControlPoint1.x, previousControlPoint1.y, previousControlPoint2.x, previousControlPoint2.y, previousPoint.x, previousPoint.y);
                    quadTo(controlPoint1.x, controlPoint1.y, pointI.x, pointI.y);
                }


                previousControlPoint1.set(controlPoint1);
                previousPoint.set(pointI);
            }
        }

    } else {
        logger.e("BezierHelper", "makeBezierCurve: error");
    }
}
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1 回答 1

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要实现您显示的链接中的效果，您需要实现 Catmull-Rom 曲线。值得庆幸的是，如果您正在使用已经可以绘制三次贝塞尔曲线的代码，那么这些实现起来很简单，因为它们是相同的曲线，只是使用不同的表示形式，并且从一种曲线转换到另一种曲线非常容易。

获取您的点列表，并为每个点创建切线，这些点与您正在查看的点之前和之后的点的线对齐。设置所有 CR 点的通用代码：

points = ...
initialpoint = points[0] - (points[1] - points[0]) // invent a 'virtual' -1th point
finalpoint = points[last] + (points[last] - points[last-1]) // invent a 'virtual' (last+1)th point
points = initialpoint + points + finalpoint
foreach ((point,i) in points) {
  try {
    p = points[i-1]
    n = points[i+1]
  } catch { continue } 
  diff = n-p 
  point.tangent = diff/2 // let's be reasonable 
}
points = points.slice(1,last-1) // throw those virtual points away

然后绘制 catmull-rom 曲线（通过将每个 CR 部分的坐标转换为贝塞尔坐标并绘制相应的贝塞尔曲线），在转换为贝塞尔坐标之前，使用页面中的滑块控制“紧密度”因子。

于 2018-05-05T08:23:35.787 回答