java - 我如何获得数组来保存所玩游戏的数量并为每个玩家运行它？

Question

我有一个作业，其中两个用户要输入他们的名字，然后获取他们玩的游戏数量，然后获取他们玩的每场游戏的分数，将分数放入一个数组(或两个数组？？？)中，然后比较两个数组和分数以查看谁赢得了每场比赛或他们是否平局，然后显示结果。我不相信我不需要对它进行排序或搜索。

我的老师说获取分数必须是一个 void 方法，但我不确定如何从这个 void 方法中将值获取到每个玩家的数组中，以便我可以比较分数值。

    public class Lab9 {
public static void inputScores(int[] array, String name)  {
    String inputScores = "";
    for (int i = 1; i < array.length; i++) {
        inputScores = JOptionPane.showInputDialog(name + " enter your score for game " + i);
        array[i] = Integer.parseInt(inputScores);
    }
}

public static void main(String[] args) {

    int numberOfGames = getPositiveIntOrQuit("How many games were played?",  "Lab 9 (by Jarvis),");



       String  name = getStringOrQuit("Player 1-What is your name?", "Lab 9 (by Jarvis");

       String name1 = getStringOrQuit(" Player 2 - What is your name?","Lab 9(by Jarvis");


        int score = getNonNegativeIntOrQuit(name  +  "  enter your score for game" ,  "Lab 9 (by Jarvis)");



        }

        public static String getStringOrQuit(String prompt, String title) {
            String userInputString;

                userInputString = JOptionPane.showInputDialog(null, prompt, title, JOptionPane.QUESTION_MESSAGE);
                // Did user hit Cancel or OK with nothing typed?
                if (userInputString == null || userInputString.trim().equals("")) {
                    JOptionPane.showMessageDialog(null,"No input, so program will terminate now.");
                    System.exit(0);
                }

            return userInputString;
        }  // getStringOrQuit



        public static int getPositiveIntOrQuit(String prompt, String title) {
            String userInputString;
            int userInputInt = 0;

            do {
                userInputString = JOptionPane.showInputDialog(null, prompt, title, JOptionPane.QUESTION_MESSAGE);
                // Did user hit Cancel or OK with nothing typed?
                if (userInputString == null || userInputString.trim().equals("")) {
                    JOptionPane.showMessageDialog(null,"No input, so program will terminate now.");
                    System.exit(0);
                }
                else
                try {
                    userInputInt = Integer.parseInt(userInputString);    // This line might throw an exception.
                    // Ok, if conversion in above line worked, check if input is a positive integer.
                    if (userInputInt < 1)
                        JOptionPane.showMessageDialog(null,"Bad input value. It must be a positive integer.",
                                "Input error", JOptionPane.ERROR_MESSAGE);
                }
                catch (NumberFormatException exc) {
                    JOptionPane.showMessageDialog(null,"Bad input value. It must be a positive integer.",
                                        "Input error", JOptionPane.ERROR_MESSAGE);
                }

            } while (userInputInt < 1);

            return userInputInt;
        }  // getPositiveIntOrQuit



        public static int getNonNegativeIntOrQuit(String prompt, String title) {
            String userInputString;
            int userInputInt = -1;

            do {
                userInputString = JOptionPane.showInputDialog(null, prompt, title, JOptionPane.QUESTION_MESSAGE);
                // Did user hit Cancel or OK with nothing typed?
                if (userInputString == null || userInputString.trim().equals("")) {
                    JOptionPane.showMessageDialog(null,"No input, so program will terminate now.");
                    System.exit(0);
                }
                try {
                    userInputInt = Integer.parseInt(userInputString);    // This line might throw an exception.
                    // Ok, if conversion in above line worked, check if input is a positive integer.
                    if (userInputInt < 0)
                        JOptionPane.showMessageDialog(null,"Bad input value. It must be a non-negative integer.",
                                "Input error", JOptionPane.ERROR_MESSAGE);
                }
                catch (NumberFormatException exc) {
                    JOptionPane.showMessageDialog(null,"Bad input value. It must be a non-negative integer.",
                                        "Input error", JOptionPane.ERROR_MESSAGE);
                }

            } while (userInputInt < 0);

            return userInputInt;



}

            }

Answer 1

我已经写了一些代码，将其作为我解决您问题的方法。我将创建一个 HashMap (或两个，每个玩家一个)并将游戏编号映射到分数。然后，您可以使用此 HashMap 轻松比较分数

示例(这是我很快就完成的粗略代码):

public class random {

private static Scanner keyboard = new Scanner(System.in);
private static HashMap<Integer, String> gameMap = new HashMap<Integer, String>();

public static void mapScores(){
    System.out.println("How many games have been played?");
    int numOfGames = keyboard.nextInt();

    String score = "";

    for(int i = 0; i < numOfGames; i++){
        System.out.println("Please enter the score for game: " + i);
        score = keyboard.nextLine();
        gameMap.put(i, score);
    }


}

}

编辑:

好吧，那么我建议静态声明你的整数数组。方便访问。你说的是一个保存玩家分数和玩家姓名的数组，这就是 id 建议使用 HashMap 的原因，因为数组无法存储名称。我将演示:

public class random {

private static Scanner keyboard = new Scanner(System.in);
private static HashMap<String, int[]> gameMap = new HashMap<String, int[]>();
private static int[] hello = new int[10];

public static void mapScores(int[] array, String name){
     String inputScores = "";
        for (int i = 1; i < array.length; i++) {
            inputScores = JOptionPane.showInputDialog(name + " enter your score for game " + i);
            array[i] = Integer.parseInt(inputScores);
        }
}

public static void main(String[] args) {

    random.gameMap.put("name", new int[10]);
    random.mapScores(gameMap.get("name"), "name");

    for(int i = 0; i < gameMap.get("name").length; i++){
        System.out.println(gameMap.get("name")[i]);
    }
}

}

显然我没有花时间正确创建对象，但希望你能看到这个想法，尽管代码很糟糕。希望这有帮助!