Question

我正在尝试编写一个程序，该程序是一个生成随机字母的游戏，用户选择随机生成的字母是否是用户名的一部分。我还没有开始第二个算法，我想专注于第一个算法。

更新：嘿，所以我跟着你们的建议，我能够让算法1工作。现在我完成算法1的唯一办法就是有一个循环来检查随机生成的字母是否在wrongLetters arrayList中。如果是，那么它将生成另一个随机字母，如果它不是arrayList的一部分，它会询问用户是否是下一个字母。 wrongLetters arraylist应该包含几个不是用户名中下一个字母的字符串（字母）。一旦正确猜到了一封信，就会清除错误的字母arraylist并在猜到下一个字母后添加新的字符串。

我尝试使用while(actualLetter != wrongLetters);使用do / while循环，但我一直收到有关“不兼容的操作数类型”的错误。我怎样才能解决这个问题？再次感谢。

再次感谢！

/*
 * PSEUDOCODE:
 * -Ask user to choose an option (algorithm 1, algorithm 2, quit)
 * -if user selects algorithm 1, generate a random number to pass off as unicode(char) and then     change char to string
 * -ask user whether generated char is part of their name
* -if yes, then add  (actualLetter) string to arrayList(name).
 * -if no, then add (actualLetter) string to arrayList(wrongLetters).
 * reask user to choose an option (algorithm 1, algorithm 2, quit)
 * whether user chooses algorithm 1 or 2, after generating random char, check char to make sure its not in the arrayList(wrongLetters)
 * if char is in the wrongLetters arrayList, then generate another random char until a char that is not in that array generates
 * 
 * 
 */


import javax.swing.JOptionPane;

import java.util.ArrayList;

public class Lab_1 {

public static void main(String[] args) {
    int n;
    ArrayList<String> name = new ArrayList<String>();
    ArrayList<String> wrongLetters = new ArrayList<String>();
    String result = "";
    int p = 1;
    Object[] options2 = {"No", "Yes"};


    do{
        do{
            Object[] options = {"Quit", "Algorithm 2", "Algorithm 1"};

            n = JOptionPane.showOptionDialog(null, "Please choose an option.", "A Question", JOptionPane.YES_NO_CANCEL_OPTION, JOptionPane.QUESTION_MESSAGE, null, options, options[0]);

            if (n != 0){
                if (n == 1){
                    JOptionPane.showMessageDialog(null, "You have chosen algorithm 2.");
                }
                else if (n == 2){
                    algorithm1(name, wrongLetters);
                    p = JOptionPane.showOptionDialog(null, "Are there any letters remaining in your name?", "A Question", JOptionPane.YES_NO_OPTION, JOptionPane.QUESTION_MESSAGE, null, options2, options2[0]);

                    //if the answer is no
                    if(p == 0){
                    for(String str: name){
                           result=result + str;
                        }
                    JOptionPane.showMessageDialog(null, "Your name is " + result);
                        }
                    }
                }
            else {
            JOptionPane.showMessageDialog(null, "Thanks for playing, " + result + "!");
            }
        }while(p == 1);
    }while (n != 0);
}

public static void algorithm1(ArrayList<String> name, ArrayList<String> wrongLetters){
    String actualLetter = "";
    do{
    int unicode = (int) (Math.random() * 25) + 65; //comes up with random number to use as a char
    char aLetter = (char) unicode;
    actualLetter = Character.toString(aLetter);
    Object[] options = {"No", "Yes"};
    int o;
    o = JOptionPane.showOptionDialog(null, "Is this the next letter in your name: " + aLetter, "A Question", JOptionPane.YES_NO_OPTION, JOptionPane.QUESTION_MESSAGE, null, options, options[0]);
    //if you choose yes
    if(o == 1){
        name.add(actualLetter);
        }
    //if you choose no
    else if (o == 0){
        wrongLetters.add(actualLetter);
    }
    }while(actualLetter != wrongLetters);
}

}

Answer 1

 JOptionPane.showMessageDialog(null, name);

这句话如果错了，就不能在名字中显示名字，因为name是 list 的变量，而不是 String 。你可以这样做：

String result="";
for(String str: name){
   result=result+";"+str;
}
JOptionPane.showMessageDialog(null, result);

Answer 2

您将字母存储在algorithm1中作为局部变量的ArrayLists中。一旦到达该功能的末尾，您将丢失两个列表以及之前存储在其中的任何值。

我不知道这是否是一个优雅的解决方案，但您可以在main方法中声明两个ArrayLists，然后将它们传递给您的算法方法。

您需要更改algorithm1的方法签名，如下所示：

public static void algorithm1(ArrayList<String> name, ArrayList<String> wrongLetters)

然后将您对algorithm1的调用更改为此，假设您分别为ArrayLists name和wrongLetters命名：

algorithm1(name, wrongLetters);

如何通过循环将字符串添加到字符串数组列表中

2 个答案: