显示返回的值，该值链接到其他两个信息数组

Question

//有没有更好的方法来显示返回的信息，而不必写我在S.O.p中写的所有内容。目标是返回值，然后将其链接到参赛者的姓名和时间。谢谢你的帮助。//

public class array_Test
{
    public static void main(String args[])
    {
        String[] names={"Bob", "Tj", "Aj"};
        double [] times={9.9,9.8,10.0};
        double x=findLargest(times);
        System.out.print("The slowest racer was " + names[1] + " with a time of " 
        + times[1] + " at index position" + x);
}
    public static double findLargest(double[] times)
    {
        int indexOfLargest=0;
        double largest=times[0];
        for(int i=0; i<times.length; i++)
        {
            if(times[i]>largest)
            {
                largest=times[i];
                indexOfLargest=i;
            }
        }
        return indexOfLargest;                      
    }}

Answer 1

为什么不使用类？你可以将参赛者的姓名和时间与以下内容合并：

public class Racer {
 private String name;
 private double time;
 private int position;
 public Person(String name, double time){
 this.name = name;
 this.time = time;
 this.position =-1;
   }
 public String getName() {
   return name;
   }
 public double getTime() {
   return time;
   }
 public int getPosition() {
   return position;
   }
 public void setPosition(int position) {
   this.position = position
   }
 @Override
 public String toString() {
   return "The slowest racer was " + getName + " with a time of " 
    + getTime + " at index position" + getPosition);
   }        
}

这个类允许您管理easlier您的"Racer"和函数toString（）的覆盖允许您自定义printl（）。

您现在可以更改您的代码：

public class array_Test
{
    public static void main(String args[])
    {
       Racer[] people ={new Person("Bob",9.9),new Person("Tj",9.8),new Person("Aj",10.0)};
        double x=findLargest(times);
        System.out.print(Racer[x].toString());
}
    public static double findLargest(double[] times)
    {
        int indexOfLargest=0;
        double largest=people[0].getTime();
        for(int i=0; i<people.length; i++)
        {
            if(people[i].getTime()>largest)
            {
                largest=people[i].getTime();
                indexOfLargest=i;
            }
        }
        people[indexOfLargest].setPosition(indexOfLargest);
        return indexOfLargest;                      
    }}