我正在尝试创建一个for循环,它将遍历9个不同的点集,并在我的主驱动程序中与另外两个类执行它们。我现在非常困难,无法弄清楚如何在for循环再次运行之前设置第二个(pointC2)。
坐标是:
X:{{86.92, 70.93, 97.74, 30.90, 51.66, 0.83, 55.91, 32.92, 65.26, 83.90},
Y:{2.47, 27.81, 34.36, 35.14, 31.70,21.77, 66.62, 75.23, 72.53, 4.710}};
带
(86.92, 2.47) = (x1,y1)
(70.93, 27.81) = (x2,y2)
我尝试将其设置为多维循环,但我无法获得正确的计数器。
分配的基础是将笛卡尔坐标转换为极坐标,然后找到从(x1,y1)
到(x2,y2)
到(x3,y3)
的距离等。
这是我需要在for循环后执行的代码:
//calls point1 from Cartesian
Cartesian pointC1 = new Cartesian(x1, y1);
//calls point2 from Cartesian
Cartesian pointC2 = new Cartesian(x2, y2);
double answer1 = Cartesian.distance(pointC1, pointC2);
//prints out point1 and point2
System.out.println("Point 1 = " + pointC2 + " Point 2 = " + pointC1);
//prints out sum
System.out.println("Distance: " + answer1);
然后重新开始,但这次是从(x2,y2)
和(x3, y3)
答案 0 :(得分:1)
我可以告诉你一个程序而不是编码,但你必须先研究arrays
!
定义,
array X: {{86.92, 70.93, 97.74, 30.90, 51.66, 0.83, 55.91, 32.92, 65.26, 83.90},
array Y: {2.47, 27.81, 34.36, 35.14, 31.70, 21.77, 66.62, 75.23, 72.53, 4.710}};
然后,
begin for loop from i = 0 to array size -1
x1 = array X [i];
y1 = array Y [i];
// rest of your code goes here
// ...
end for loop
答案 1 :(得分:0)
double[] x = {86.92, 70.93, 97.74, 30.90, 51.66, 0.83, 55.91, 32.92, 65.26, 83.90};
double[] y = {2.47, 27.81, 34.36, 35.14, 31.70, 21.77, 66.62, 75.23, 72.53, 4.710};
Cartesian[] pointC = new Cartesian[Math.min(x.length, y.length)];
double[] distance = new double[pointC.length - 1];
for(int i = 0; i < pointC.length; i++){
pointC[i] = new Cartesian(x[i], y[i]);
}
for(int i = 0; i < distance.length; i++){
distance[i] = Cartesian.distance(pointC[i], pointC[i+1]);
}