java - 不要重复字符 - Java

Question

如果输入的字符已被使用，我试图让程序打印出错误。如果您检查 else If 语句，从技术上讲它会运行，但我不希望它每次都打印。仅当该角色已在另一轮中被选择时。

while (!userCorrect) {
    Scanner input = new Scanner(System.in);
    System.out.print("Guessing (round " + roundNumber + "): Choosing your letter from a-z: ");
    String letters = input.nextLine();
    createString(letters.length());
    if (letters.length () > 1) { //code will print an error if the user selects more than 1 character
        System.out.println("You should not enter more than 1 character");
    }
    else if (letters.length()==letters.length()) { //print error if the character has been chosen already
        System.out.println(letters + " has been chosen before, try again");
    }
    else {
        System.out.println("end (round " + roundNumber + ")");
        roundNumber++; 
    }
}

我得到的最接近的是letters.length() == letters.length();但我知道这样做总是会打印错误。您认为我可以在其中放入一个参数，使其跳过第一轮并仅在第二轮开始时才开始查找吗？

Answer 1

我知道你已经有了答案，但我想指出一些事情

• 不要每次都在循环中创建新的 Scanner 对象
• 使用继续避免嵌套if-else
• 有多种方法可以存储字符，例如使用 List 或 Set 集合或 StringBuffer 或简单的 String。
• 您的 boolean 变量的名称是...
• 我忽略了 createString 方法并跳过了 userCorrect，因为他们不清楚如何使用

(我也结束了自己的循环退出条件，因为只是为了有一个)

Scanner input = new Scanner(System.in);
int roundNumber = 1;
boolean isUserCorrect = false;

String selected = "";

while (true) {
    System.out.print("Guessing (round " + roundNumber + "): Choosing your letter from a-z: ");
    String letters = input.nextLine();

    if (letters.length () > 1) { //code will print an error if the user selects more than 1 character
        System.out.println("You should not enter more than 1 character");
        continue;
    }

    if (selected.contains(letters)) {
        System.out.println(letters + " has been chosen before, try again");
        continue;
    }

    selected = selected.concat(letters);
    System.out.println("end (round " + roundNumber + ")");
    roundNumber++;
    if (roundNumber > 5) {
        System.out.println("Selected characters: " + selected);
        break;
    }
}
input.close();