根据四个点坐标求两个直线的交叉点坐标（Java&JavaScript）JavaJavaScript

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我是靠谱客的博主整齐洋葱，最近开发中收集的这篇文章主要介绍根据四个点坐标求两个直线的交叉点坐标（Java&JavaScript）JavaJavaScript，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

Java

	//计算两条直线的交点
    //输入四个点，前两个点确定线段1，后两个点确定线段2
    //输出交点坐标
    public static void main(String args[]){
        double[] point1= new double[]{0,0};
        double[] point2 = new double[]{2,2};

        double[] point3 = new double[]{0,2};
        double[] point4 = new double[]{2,0};
        double[] node = cross(point1,point2,point3,point4);
        if(node==null)System.out.println(false);
        else System.out.println(node[0]+","+node[1]);
        HashSet<String> set = new HashSet<>();
        set.add(node[0]+":"+node[1]);
        for(String item:set){
            System.out.println(item);
        }
    }

    public static double[] cross(double[] point1,double[] point2,double[] point3,double[] point4){
        if((point2[1]-point1[1])*(point4[0]-point3[0])==(point4[1]-point3[1])*(point2[0]-point1[0]))return null;
        if(point2[0]-point1[0]==0){
            double k2 = (point4[1]-point3[1])/(point4[0]-point3[0]);
            double b2 = point3[1]-point3[0]*k2;
            double cross_x = point1[0];
            double cross_y = k2*cross_x+b2;
            return new double[]{cross_x,cross_y};
        }
        if(point4[0]-point3[0]==0){
            double k1 = (point2[1]-point1[1])/(point2[0]-point1[0]);
            double b1 = point1[1]-point1[0]*k1;
            double cross_x = point3[0];
            double cross_y = k1*cross_x+b1;
            return new double[]{cross_x,cross_y};
        }
        double k1 = (point2[1]-point1[1])/(point2[0]-point1[0]);
        double b1 = point1[1]-point1[0]*k1;
        double k2 = (point4[1]-point3[1])/(point4[0]-point3[0]);
        double b2 = point3[1]-point3[0]*k2;
        double cross_x = (b1-b2)/(k2-k1);
        double cross_y = k1*cross_x+b1;
        return new double[]{cross_x,cross_y};
    }

JavaScript

//四个坐标 两条直线 求交叉点
function segmentsIntr(a, b, c, d){
    /** 1 解线性方程组, 求线段交点. **/
// 如果分母为0 则平行或共线, 不相交
    var denominator = (b[1] - a[1])*(d[0] - c[0]) - (a[0] - b[0])*(c[1] - d[1]);
    if (denominator==0) {
        return false;
    }
// 线段所在直线的交点坐标 (x , y)
    var x = ( (b[0] - a[0]) * (d[0] - c[0]) * (c[1] - a[1])
        + (b[1] - a[1]) * (d[0] - c[0]) * a[0]
        - (d[1] - c[1]) * (b[0] - a[0]) * c[0] ) / denominator ;
    var y = -( (b[1] - a[1]) * (d[1] - c[1]) * (c[0] - a[0])
        + (b[0] - a[0]) * (d[1] - c[1]) * a[1]
        - (d[0] - c[0]) * (b[1] - a[1]) * c[1] ) / denominator;
    /** 2 判断交点是否在两条线段上 **/
    if (
        // 交点在线段1上
        (x - a[0]) * (x - b[0]) <= 0 && (y - a[1]) * (y - b[1]) <= 0
        // 且交点也在线段2上
        && (x - c[0]) * (x - d[0]) <= 0 && (y - c[1]) * (y - d[1]) <= 0
    ){
        // 返回交点p
        return [x,y]
    }
    //否则不相交
    return false
}