本文主要是介绍贝塞尔曲线 java，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
参考文章：
理解贝塞尔曲线https://blog.csdn.net/weixin_42301220/article/details/125167672
代码实现参考
https://blog.csdn.net/yinhun2012/article/details/118653732
贝塞尔 一二三阶java代码实现,N阶段可以通过降阶递归实现
public class Test extends JPanel {@Overridepublic void paintComponent(Graphics g) {super.paintComponent(g);Graphics2D g2 = (Graphics2D)g;//画图测试贝塞尔曲线ArrayList<T> list = test1B();for (int i = 0; i < list.size(); i++) {//画线的2个点一样即画点g2.drawLine((int)list.get(i).p.x,(int)list.get(i).p.y,(int)list.get(i).p.x,(int)list.get(i).p.y);}list = test2B();for (int i = 0; i < list.size(); i++) {//画线的2个点一样即画点g2.drawLine((int)list.get(i).p.x,(int)list.get(i).p.y,(int)list.get(i).p.x,(int)list.get(i).p.y);if(i%100==0){//展示部分切线g2.drawLine((int)list.get(i).a1.x,(int)list.get(i).a1.y,(int)list.get(i).a2.x,(int)list.get(i).a2.y);}}list = test3B();for (int i = 0; i < list.size(); i++) {//画线的2个点一样即画点g2.drawLine((int)list.get(i).p.x,(int)list.get(i).p.y,(int)list.get(i).p.x,(int)list.get(i).p.y);if(i%100==0){//展示部分切线g2.drawLine((int)list.get(i).a1.x,(int)list.get(i).a1.y,(int)list.get(i).a2.x,(int)list.get(i).a2.y);}}}static class T{Point p;Point a1;//p的切线起始点Point a2;//p的切线结束点}//测试贝塞尔曲线//1阶 2个点static T get1Bse(Point p1,Point p2,double t){T tp = new T();tp.p = new Point();tp.a1 = p1;tp.a2 = p2;tp.p.x = (1-t)*p1.x + t*p2.x;tp.p.y = (1-t)*p1.y + t*p2.y;return tp;}static ArrayList<T> test1B(){ArrayList<T> list = new ArrayList<>();for (int i = 0; i <= 1000; i++) {T tt = get1Bse(new Point(100,100,""),new Point(200,200,""),i*0.001);list.add(tt);}return list;}//2阶 3个点static T get2Bse(Point p1,Point p2,Point p3,double t){T tp = new T();tp.p = new Point();tp.a1 = new Point();tp.a2 = new Point();tp.a1.x = (1-t)*p1.x + t*p2.x;tp.a1.y = (1-t)*p1.y + t*p2.y;tp.a2.x = (1-t)*p2.x + t*p3.x;tp.a2.y = (1-t)*p2.y + t*p3.y;tp.p.x = (1-t)*(1-t)*p1.x + 2*t*(1-t)*p2.x + t*t*p3.x;tp.p.y = (1-t)*(1-t)*p1.y + 2*t*(1-t)*p2.y + t*t*p3.y;return tp;}static ArrayList<T> test2B(){ArrayList<T> list = new ArrayList<>();for (int i = 0; i <= 1000; i++) {T tt = get2Bse(new Point(100,300,""),new Point(200,300,""),new Point(200,400,""),i*0.001);list.add(tt);}return list;}//3阶 4个点static T get3Bse(Point p1,Point p2,Point p3,Point p4,double t){Point p5 = new Point();Point p6 = new Point();Point p7 = new Point();p5.x = (1-t)*p1.x + t*p2.x;p5.y = (1-t)*p1.y + t*p2.y;p6.x = (1-t)*p2.x + t*p3.x;p6.y = (1-t)*p2.y + t*p3.y;p7.x = (1-t)*p3.x + t*p4.x;p7.y = (1-t)*p3.y + t*p4.y;T tp = new T();tp.p = new Point();tp.a1 = new Point();tp.a2 = new Point();tp.a1.x = (1-t)*p5.x + t*p6.x;tp.a1.y = (1-t)*p5.y + t*p6.y;tp.a2.x = (1-t)*p6.x + t*p7.x;tp.a2.y = (1-t)*p6.y + t*p7.y;tp.p.x = (1-t)*(1-t)*p5.x + 2*t*(1-t)*p6.x + t*t*p7.x;tp.p.y = (1-t)*(1-t)*p5.y + 2*t*(1-t)*p6.y + t*t*p7.y;return tp;}static ArrayList<T> test3B(){ArrayList<T> list = new ArrayList<>();for (int i = 0; i <= 1000; i++) {T tt = get3Bse(new Point(100,500,""),new Point(200,600,""),new Point(300,450,""),new Point(400,300,""),i*0.001);list.add(tt);}return list;}/*n阶的实现: todo
因为贝塞尔曲线的规则可以让n个顶点的问题降维到n-1的顶点的计算，那么我们递归套娃，是不是将n个顶点的问题变成由最基本的3个顶点(2阶)计算规则来实现呢？
例如：5个顶点的问题，先降到成4个顶点，再降维到3个顶点*/
}
效果图
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