Map<String, Integer> cityAppSum = new HashMap<>();cityAppSum.get(city)==null?cityAppSum.put(city,1): cityAppSum.put(city,Integer.parseInt(cityAppSum.get(city)+"")+=1);这段代码中有什么问题请指出并修正

在这段代码中，存在以下问题：

1. 在使用 `cityAppSum.put(city,1)` 语句时，没有考虑 `city` 是否已经存在于 `cityAppSum` 中。如果 `city` 已经存在，这将覆盖之前的值。

2. 在使用 `cityAppSum.put(city,Integer.parseInt(cityAppSum.get(city)+"")+=1)` 语句时，尝试将整数值增加 1 并更新到 `cityAppSum` 中。但是，`Integer.parseInt()` 方法不支持对表达式的解析和计算。

这是修正后的代码：

Map<String, Integer> cityAppSum = new HashMap<>();
int count = cityAppSum.getOrDefault(city, 0);
cityAppSum.put(city, count + 1);

这段修正后的代码使用了 `getOrDefault()` 方法来获取 `city` 对应的值，如果该键不存在，则返回默认值0。然后，将旧值加1并更新到 `cityAppSum` 中。这样可以正确地计算每个城市的应用数量。

相关问题

如何在Java的Customer类中实现一个HashMap来存储Ticket对象？

为了在Java的`Customer`类中使用`HashMap`来存储`Ticket`对象，你可以按照以下步骤进行：

### 1. 导入必要的包
首先，确保导入`java.util.HashMap`和`java.util.Map`包。

import java.util.HashMap;
import java.util.Map;

### 2. 定义`Customer`类
在`Customer`类中定义一个`HashMap`来存储每种票的数量。

public class Customer {
    private String firstName;
    private String surname;
    private Map<Ticket, Integer> tickets;

    // 构造方法
    public Customer(String firstName, String surname) {
        this.firstName = firstName;
        this.surname = surname;
        this.tickets = new HashMap<>();
    }

    // 获取客户的名字
    public String getFirstName() {
        return firstName;
    }

    // 设置客户的名字
    public void setFirstName(String firstName) {
        this.firstName = firstName;
    }

    // 获取客户的姓氏
    public String getSurname() {
        return surname;
    }

    // 设置客户的姓氏
    public void setSurname(String surname) {
        this.surname = surname;
    }

    // 添加票
    public void addTicket(Ticket ticket, int quantity) {
        if (tickets.containsKey(ticket)) {
            int currentQuantity = tickets.get(ticket);
            tickets.put(ticket, currentQuantity + quantity);
        } else {
            tickets.put(ticket, quantity);
        }
    }

    // 移除票
    public boolean removeTicket(Ticket ticket, int quantity) {
        if (tickets.containsKey(ticket)) {
            int currentQuantity = tickets.get(ticket);
            if (currentQuantity >= quantity) {
                if (currentQuantity == quantity) {
                    tickets.remove(ticket);
                } else {
                    tickets.put(ticket, currentQuantity - quantity);
                }
                return true;
            }
        }
        return false;
    }

    // 获取所有票的信息
    public Map<Ticket, Integer> getTickets() {
        return tickets;
    }

    // 计算总成本（包括折扣）
    public double calculateTotalCost() {
        double totalCost = 0.0;
        int totalTickets = 0;

        for (Map.Entry<Ticket, Integer> entry : tickets.entrySet()) {
            Ticket ticket = entry.getKey();
            int quantity = entry.getValue();
            totalCost += ticket.getPrice() * quantity;
            totalTickets += quantity;
        }

        // 应用折扣
        if (totalTickets >= 26) {
            totalCost *= 0.75; // 25% 折扣
        } else if (totalTickets >= 11) {
            totalCost *= 0.85; // 15% 折扣
        } else if (totalTickets >= 6) {
            totalCost *= 0.90; // 10% 折扣
        }

        return totalCost;
    }

    // 计算原始总成本（不包括折扣）
    public double calculateOriginalTotalCost() {
        double totalCost = 0.0;

        for (Map.Entry<Ticket, Integer> entry : tickets.entrySet()) {
            Ticket ticket = entry.getKey();
            int quantity = entry.getValue();
            totalCost += ticket.getPrice() * quantity;
        }

        return totalCost;
    }
}

### 3. `Ticket`类的定义
假设你已经有一个`Ticket`类，它至少包含票价和名称属性。

public class Ticket implements Comparable<Ticket> {
    private String name;
    private double price;

    public Ticket(String name, double price) {
        this.name = name;
        this.price = price;
    }

    public String getName() {
        return name;
    }

    public double getPrice() {
        return price;
    }

    @Override
    public int compareTo(Ticket other) {
        return this.name.compareTo(other.name);
    }

    @Override
    public boolean equals(Object obj) {
        if (this == obj) return true;
        if (obj == null || getClass() != obj.getClass()) return false;
        Ticket ticket = (Ticket) obj;
        return name.equals(ticket.name);
    }

    @Override
    public int hashCode() {
        return name.hashCode();
    }
}

### 4. 使用示例
以下是如何在`MainProgram`类中使用`Customer`和`Ticket`类的示例。

public class MainProgram {
    public static void main(String[] args) {
        // 创建一些票
        Ticket ticket1 = new Ticket("Magpie Line", 5.59);
        Ticket ticket2 = new Ticket("Cross City", 2.50);

        // 创建一个客户
        Customer customer = new Customer("Pam", "Royal");

        // 添加票
        customer.addTicket(ticket1, 5);
        customer.addTicket(ticket2, 10);

        // 显示客户信息
        System.out.println("Customer: " + customer.getFirstName() + " " + customer.getSurname());
        System.out.println("Original Total Cost: " + customer.calculateOriginalTotalCost());
        System.out.println("Discounted Total Cost: " + customer.calculateTotalCost());

        // 移除票
        customer.removeTicket(ticket1, 2);

        // 再次显示客户信息
        System.out.println("After removing tickets:");
        System.out.println("Original Total Cost: " + customer.calculateOriginalTotalCost());
        System.out.println("Discounted Total Cost: " + customer.calculateTotalCost());
    }
}

通过以上步骤，你可以在`Customer`类中使用`HashMap`来存储和管理不同类型的`Ticket`对象及其数量，并计算相关的费用和折扣。

写一段Java实现Input In the first line you are given the number n of billionaires (1 ≤ n ≤ 10000). The following n lines contain information about these people: their names, cities where they were staying at the beginning of the period, and their fortunes. In the next line you are given the number m of days in the period for which you have the information (1 ≤ m ≤ 50000) and the number k of travels of the billionaires (0 ≤ k ≤ 50000). The following k lines contain the list of travels in the following format: the number of the day (from 1 to m−1), the name of the person, and the city of destination. You may assume that billionaires depart late at night and arrive to the destination city on the next day's morning. They cannot make more than one travel each day. The numbers of days in the list are not decreasing. All names of people and cities consist of at most 20 English letters; you must take into consideration the case of the symbols. The fortunes are in the range from 1 to 100 billions (one billion is a thousand million). Output In each line of the output give the name of a city and, after a space, the number of days during which this city was the first with respect to the sum of fortunes of the billionaires staying there. Leave out those cities for which there were no such days. The cities must be sorted alphabetically (with the usual symbol order: ABC...Zabc...z).

以下是Java实现代码：

import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        Map<String, Integer> cityDays = new TreeMap<>();
        Map<String, Integer> cityFortunes = new HashMap<>();
        Map<String, String> billionaireCities = new HashMap<>();

        for (int i = 0; i < n; i++) {
            String name = scanner.next();
            String city = scanner.next();
            int fortune = scanner.nextInt();
            billionaireCities.put(name, city);
            cityFortunes.put(city, cityFortunes.getOrDefault(city, 0) + fortune);
        }

        int m = scanner.nextInt();
        int k = scanner.nextInt();
        for (int i = 0; i < k; i++) {
            int day = scanner.nextInt();
            String name = scanner.next();
            String destCity = scanner.next();
            String origCity = billionaireCities.get(name);
            int fortune = cityFortunes.get(origCity);
            if (cityDays.get(origCity) == null) {
                cityDays.put(origCity, 0);
            }
            if (cityDays.get(destCity) == null) {
                cityDays.put(destCity, 0);
            }
            cityDays.put(origCity, cityDays.get(origCity) + day - cityDays.get(destCity));
            cityDays.put(destCity, day);
            cityFortunes.put(origCity, fortune - cityFortunes.getOrDefault(origCity, 0));
            cityFortunes.put(destCity, cityFortunes.getOrDefault(destCity, 0) + fortune);
            billionaireCities.put(name, destCity);
        }

        List<String> result = new ArrayList<>();
        for (Map.Entry<String, Integer> entry : cityDays.entrySet()) {
            String city = entry.getKey();
            int days = entry.getValue();
            if (days > 0) {
                result.add(city + " " + days);
            }
        }

        for (String s : result) {
            System.out.println(s);
        }
    }
}

解释一下代码：

首先读取输入的数据，使用Map<String, Integer> cityDays记录每个城市的首富天数，Map<String, Integer> cityFortunes记录每个城市的首富总财富，Map<String, String> billionaireCities记录每个亿万富翁所在的城市。

接着，遍历每个亿万富翁，将其所在城市的首富总财富加入到cityFortunes中。然后，读取每个亿万富翁的旅行记录，更新cityDays、cityFortunes和billionaireCities。对于每条旅行记录，先计算出旅行前该亿万富翁所在城市的首富天数，然后将其更新为旅行前最后一次到达该城市的天数减去旅行天数，将目的地城市的首富天数更新为旅行天数，更新原城市和目的地城市的首富总财富，并将该亿万富翁所在城市更新为目的地城市。

最后，遍历cityDays，将首富天数大于0的城市加入到结果列表中，并按照字典序排序输出。