地图算法（四）：判断当前点是不是在圆内

圆面积判断器实现

最新推荐文章于 2024-07-05 03:24:26 发布

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最新推荐文章于 2024-07-05 03:24:26 发布

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本文链接：https://blog.youkuaiyun.com/liu3364575/article/details/78393292

本文介绍了一个名为CircleAreaJudge的类的实现，该类用于判断一个坐标点是否位于指定圆的内部。通过构造函数初始化圆心坐标及半径，并提供了一个方法isInArea来接收坐标点，使用RuleUtil类中的computeLength方法计算距离并判断该点是否在圆内。

public class CircleAreaJudge {
    private double cx = 0.0D;
    private double cy = 0.0D;
    private double r = 0.0D;

    public CircleAreaJudge(double x, double y, double r) {
        try {
            this.cx = x;
            this.cy = y;
            this.r = r;
        } catch (NumberFormatException localNumberFormatException) {
        }
    }

    public boolean isInArea(double x, double y) {
        return isInArea(x, y);
    }


    public boolean isInArea(String x, String y) {
        double dis = RuleUtil.computeLength(Double.parseDouble(x), Double.parseDouble(y), this.cx, this.cy);
        if (dis < this.r) {
            return true;
        }
        return false;
    }

    public static void main(String[] args) {
        CircleAreaJudge circleAreaJudge=new CircleAreaJudge(2.0, 2.0, 1);
        System.out.println(circleAreaJudge.isInArea("1","1"));
    }
}

RuleUtil.computeLength

    public static double computeLength(double dwStartX, double dwStartY, double dwEndX, double dwEndY) {
        double PI = 3.1415926535898D;

        double latRadians1 = dwStartY * (PI / 180.0D);
        double latRadians2 = dwEndY * (PI / 180.0D);
        double latRadians = latRadians1 - latRadians2;
        double lngRadians = dwStartX * (PI / 180.0D) - dwEndX * (PI / 180.0D);
        double f = 2.0D * Math.asin(Math.sqrt(Math.pow(Math.sin(latRadians / 2.0D), 2.0D)
                + Math.cos(latRadians1) * Math.cos(latRadians2) * Math.pow(Math.sin(lngRadians / 2.0D), 2.0D)));
        return f * 6378137.0D;
    }