Question

在我编写的这个程序中，我必须打印出在商店中花费最多的客户的名字。我需要帮助在数组列表中搜索花费最多的客户。

 package bestcustomer;
 import java.util.*;
/**
 *
 * @author muf15
 */
public class BestCustomer {

    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        ArrayList<Double> sales = new ArrayList<Double>();
        ArrayList<String> names = new ArrayList<String>();
        double salesAmount;
        System.out.println("Enter the sales for first customer: ");
        salesAmount = in.nextDouble();
        while(salesAmount !=0)
        {
            sales.add(salesAmount);
            System.out.println("Enter customers name");
            names.add(in.next());
            System.out.println("Enter the next sales amount, 0 to exit: ");
            salesAmount = in.nextDouble();
        }
        String bestCustomer = nameOfBestCustomer(sales, names);

    }
    public static String nameOfBestCustomer(ArrayList<Double> sales,
            ArrayList<String> customers)
    {
        String name = "";
        double maxSales;



        return name;
    }


}

Answer 1

您应该将这两个字段包含在名为Customer的类中，然后再包含

使用Collections.max();

Collections.max(yourCollection, customComparator);

Answer 2

您应该考虑让Customer成为一个类，但是这会找到您当前数据结构的名称：

  public static String nameOfBestCustomer(ArrayList<Double> sales,
        ArrayList<String> customers)
{
    String name = "";
    double maxSales = 0;
    int index = -1;

    for(int i = 0; i < sales.size(); i++) {
       if(sales.get(i) > maxSales) {
         index = i;
         maxSales = sales.get(i);
       }
    }

    if(index == -1) {
       return null; //lists are empty
    }

    return customers.get(index);
}

Answer 3

我可能有点晚了.. 我认为，如果您创建一个包含两个字段name和sale的Customer类，那么其他答案就会提到更好的设计。然后在BestCustomer中，您可以遍历客户列表，找到最高销售额并返回名称。对BestCustomer

这样的事情

private ArrayList<Customer> customers = new ArrayList<Customer>();

public BestCustomer(){
    Scanner in = new Scanner(System.in);
    double salesAmount;
    System.out.println("Enter the sales for first customer: ");
    salesAmount = in.nextDouble();
    while(salesAmount !=0)
    {
        System.out.println("Enter customers name");
        String name = in.next();
        customers.add(new Customer(name, salesAmount));
        System.out.println("Enter the next sales amount, 0 to exit: ");
        salesAmount = in.nextDouble();
    }
    String bestCustomer = nameOfBestCustomer();
    System.out.print(bestCustomer);
}

private double highestSale(){
    double highestSale = 0;
    for(Customer c: customers)
        if (c.getSales() > highestSale)
            highestSale = c.getSales();

    return highestSale;
}

public String nameOfBestCustomer(){
    for (Customer c: customers)
        if(c.matchSale(highestSale()))
            return c.getName();
    return null;
}

}

这是Customer

public class Customer {
private String name;
private double sales;



public Customer(String name, double salesAmount) {
    this.name = name;
    sales = salesAmount;
}

public boolean matchSale(double sales){
    return this.sales == sales;
}

public double getSales(){
    return sales;
}

public String getName(){
    return name;
}

}

我是初学者，所以我很确定有更有效的方法。我也使用两个getters，据我所知，这不是更好的设计..

Answer 4

在Java8中，如果像Customer这样定义mikey，则

   customers.stream()
             .max(Comparator.comparing(Customer::getSales))
             .get()
             .getName()

查找arraylist中的最大值

4 个答案: