Question

我正在尝试制作一个刽子手游戏。假设要猜的是＆＃34;灯泡。＆＃34;当用户猜出在单词中出现多次的字母时，在这种情况下是＆＃34; l，＆＃34;我想从所有位置的ArrayList中删除该字母，以便ArrayList包含＆＃34; ightbub。＆＃34;下面是我处理userGuesses的类的代码，并检查所选单词中的字母。以下代码：wordToGuess =（＆＃34;灯泡＆＃34;）

另一个问题：在此代码下方，我已经包含了输出。正如您所看到的，当我重复输入相同的字母时，它表示我已经猜到了一封信，即使该字母应该从ArrayList中删除。

import java.util.ArrayList;

public class InputChecker extends PromptUser{

static int numOfGuesses;
static String userGuess;
static String[] charList;
static boolean contains;

ArrayList<String> characters = new ArrayList<String>();

public boolean alive = true;

static WordSelector ws = new WordSelector();
static LengthReturner lr = new LengthReturner();
static PromptUser pu = new PromptUser();

static String wordToGuess = ws.setWord();
static int wordLength;

public static void finish() {
    System.out.println();
    System.out.println();
    System.out.println("Congratulations! You have finished the game.");
    System.out.println("Coming soon: GUI");
}

public boolean askUser() {
    while(alive == true) {
        boolean validGuess = false;
        while(validGuess == false) {
            userGuess = pu.getUserInput("Enter your guess here:");
            if(userGuess == null || userGuess.length() > 1) {
                validGuess = false;
                System.out.println("Enter ONE letter!");
            } else {
                validGuess = true;
            }
        }   
        wordLength = lr.setLength(wordToGuess);
        charList = wordToGuess.split("(?!^)");
        contains = false;
        for(String c : charList) {
            characters.add(c);
        }
        for (String c : characters) {
            if(userGuess == c) {
                characters.remove(c);
                contains = true;
            }
        }
        if(contains == true) {
            System.out.println("You guessed a letter!");
            numOfGuesses++;
            if(characters.isEmpty()) {
                System.out.println();
                System.out.println();
                System.out.println("Congratulations! You have successfully guessed the word, which was " + wordToGuess + ".");
                System.out.println("You took " + numOfGuesses + " guesses!");
                alive = false;
            }
        } else {
            System.out.println("Your guess, '" + userGuess + "', is not in the word!");
            numOfGuesses++;
        }
    }
    return alive;
}
}

这是我重复输入相同字母时的输出。注意它总是说我成功猜到了一封信，即使这封信应该从ArrayList中删除。不知何故，remove（）不适合我。同样，wordToGuess =（＆＃34;灯泡:)）

Enter your guess here: 
l
You guessed a letter!
Enter your guess here: 
l
You guessed a letter!
Enter your guess here: 
l
You guessed a letter!
Enter your guess here: 
l
You guessed a letter!
Enter your guess here: 
l
You guessed a letter!
Enter your guess here: 
l
You guessed a letter!
Enter your guess here: 
l
You guessed a letter!
Enter your guess here: 
l
You guessed a letter!
Enter your guess here: 
l
You guessed a letter!

Answer 1

移动线条：

    charList = wordToGuess.split("(?!^)");
    contains = false;
    for(String c : charList) {
        characters.add(c);
    }

在while循环之外的

。由于它在while循环中，它正在重新初始化＆＃34;字符＆＃34;在每次迭代中。

Answer 2

Iterator.remove()的JavaDoc（部分），

每次调用next()时，只能调用一次此方法。 迭代器的行为未指定，如果在迭代进行中修改基础集合，除了通过调用此方法之外的其他任何方式。

另外，您将void Main() { List<Music> myMusic = new List<Music> { new Music { Artist = "Mozart", Album = "Mozarts amazing album", TotalTracks = int.MaxValue, Etc = int.MinValue }, new Music { Artist = "Foo", Album = "Bar", TotalTracks = int.MaxValue, Etc = int.MinValue }, }; var mozartsMusic = myMusic.Where(music => music.Artist == "Mozart") .ToList(); mozartsMusic.ForEach(Console.WriteLine); } public class Music { public string Artist { get; set; } public string Album { get; set; } public int TotalTracks { get; set; } public int Etc { get; set; } public override string ToString() { return string.Join("\n",this.GetType().GetProperties().Select(p=>string.Format("{0} {1}", p.Name, p.GetValue(this)))); } }种类型（包括Object）平等与String（不是equals()）进行比较。你需要像

这样的东西

==

ArrayList（删除问题）

2 个答案: