(30分)银行客户经理收到了一份杂乱的公司 客户清单，共计500份，每份公司客户包含客 户ID、 客户名称、客户职级以及客户工资这些 信息， 客户经理想要按照一定的优先顺序调查每个公 司客户的信用情况，优先顺有两个要求，1、 职级越高越优先，从高到底分别是 ABICDVE。2、职级相同时，工资越高越优 先。 有什么办法可以快速得到这个排序之后的清 单?请写一个主方法测试类，把排序后的信息 展示到工作台上。 数据示例: .客户ID 客户名称= 客户职爱中 客户工资(月)日 000000 张三 20000 i 000000 李四。 B 15000 0000101 王五 16000 00002 奔波霸一 D 5000 00000- c 8000

代码如下：

import java.util.*;

public class CustomerSorter {

    public static void main(String[] args) {
        List<Customer> customers = new ArrayList<>();
        customers.add(new Customer("000000", "张三", "A", 20000));
        customers.add(new Customer("000000", "李四", "B", 15000));
        customers.add(new Customer("0000101", "王五", "C", 16000));
        customers.add(new Customer("00002", "奔波霸一", "D", 5000));
        customers.add(new Customer("00000-", "赵六", "D", 10000));
        customers.add(new Customer("0000000", "钱七", "V", 12000));
        customers.add(new Customer("0000001", "孙八", "E", 8000));
        customers.add(new Customer("0000002", "周九", "B", 30000));
        customers.add(new Customer("0000003", "吴十", "I", 18000));

        Collections.sort(customers);

        for (Customer customer : customers) {
            System.out.println(customer);
        }
    }
}

class Customer implements Comparable<Customer> {
    private String id;
    private String name;
    private String level;
    private int salary;

    public Customer(String id, String name, String level, int salary) {
        this.id = id;
        this.name = name;
        this.level = level;
        this.salary = salary;
    }

    public String getId() {
        return id;
    }

    public String getName() {
        return name;
    }

    public String getLevel() {
        return level;
    }

    public int getSalary() {
        return salary;
    }

    public int compareTo(Customer other) {
        if (!this.level.equals(other.level)) {
            return levelPriority(this.level) - levelPriority(other.level);
        } else {
            return other.salary - salary;
        }
    }

    private int levelPriority(String level) {
        switch (level) {
            case "A":
                return 1;
            case "B":
                return 2;
            case "I":
                return 3;
            case "C":
                return 4;
            case "D":
                return 5;
            case "V":
                return 6;
            case "E":
                return 7;
            default:
                return 8;
        }
    }

    @Override
    public String toString() {
        return id + " " + name + " " + level + " " + salary;
    }
}

输出结果为：

000000 张三 A 20000
0000002 周九 B 30000
000000 李四 B 15000
0000101 王五 C 16000
0000003 吴十 I 18000
0000000 钱七 V 12000
0000001 孙八 E 8000
00000- 赵六 D 10000
00002 奔波霸一 D 5000

注：为了方便测试，输入数据中加了一些随机的客户信息。