IndexOutOfBoundsException 时尝试从 ArrayList 获取字符串

Question

所以我正在尝试创建一个小型刽子手迷你游戏。不幸的是，我不断收到相同的错误。我已经尝试了两次重做游戏，但似乎仍然无法正确。我敢肯定这只是一些逻辑错误或其他什么。

我对Java还比较陌生，所以任何帮助将不胜感激。

我将程序分为三个类。一个用于 Hangman 对象，一个用于包含所述对象的数组，一个用于使用这些对象的数组。下面的三个部分是游戏的片段。

第1类：刽子手

package hangman2;
import java.util.*;
import java.io.*;
public class Hangman {
private ArrayList<String> words = new ArrayList<>();
private int diff;
public Hangman(String fileName, int difficulty) {
diff = difficulty;
//easy = 0; medium = 1; hard = 2
try {
Scanner scanFile  = new Scanner(new File(fileName));
while(scanFile.hasNext()) {
String line = scanFile.nextLine();
Scanner scanLine = new Scanner(line);
String tempWord = scanLine.next();
words.add(tempWord);
}
}
catch (FileNotFoundException e) {
System.out.println("File not foundn" + e);
}
}
//Gets
public ArrayList<String> getWords() { return words; }
public int getDifficulty() { return diff; }
//Sets
public void addWord(String word) { words.add(word); }
}

第2类：刽子手阵列

package hangman2;
import java.util.ArrayList;
import javax.swing.JOptionPane;
public class HangmanArray {
private Hangman easyHangman;
private Hangman mediumHangman;
private Hangman hardHangman;
public HangmanArray() {
easyHangman = new Hangman("easy.txt", 0);
mediumHangman = new Hangman("medium.txt", 1);
hardHangman = new Hangman("hard.txt", 2);
}
public String getRandomWord() {
Hangman workingHangman = chooseDifficulty();
int max = workingHangman.getWords().size();
int randIndex = (int) (Math.random() * max);
return workingHangman.getWords().get(randIndex);
}
private Hangman chooseDifficulty() {
int chosenDifficulty = Integer.parseInt(JOptionPane.showInputDialog("Choose difficulty leveln"
+ "1: Novicen2: Intermediaten3: Expert"));
Hangman retHangman = null;
switch(chosenDifficulty) {
case 1: retHangman = easyHangman; break;
case 2: retHangman = mediumHangman; break;
case 3: retHangman = mediumHangman; break;
}
if (retHangman == null) {
System.out.println("Chosen difficulty not within range of [1;3]nexiting");
System.exit(0);
}
return retHangman;
}
}

第3类：刽子手UI

package hangman2;
public class HangmanUI {
public static void main(String[] args) {
HangmanArray hangmanArray = new HangmanArray();
System.out.println(hangmanArray.getRandomWord());
}
}

当选择难度 2 或 3 时，错误似乎来自 HangmanArray 的第 25 行：

return workingHangman.getWords().get(randIndex);

任何帮助，不胜感激。

Answer 1

您应该检查max是否为 0，如果是，则返回null或空字符串，然后处理此特殊情况。
我用了 Random.nextInt(( 而不是int randIndex = (int) (Math.random() * max);

返回从此随机数生成器的序列中提取的介于 0(含(和指定值(不含(之间的均匀分布的伪随机 int 值。nextInt 的一般约定是，指定范围内的一个 int 值是伪随机生成并返回的。所有有界可能的 int 值都以(近似(相等的概率生成

请记住，最后一个元素的索引是size()-1

public String getRandomWord() {
Hangman workingHangman = chooseDifficulty();
int max = workingHangman.getWords().size();
if (max == 0) {
return "";
}
int randIndex = new Random().nextInt(max); // returns an integer between 0 and max-1
return workingHangman.getWords().get(randIndex);
}  

您还应该使用试用资源来确保您使用的资源在处理后被关闭。

public Hangman(String fileName, int difficulty) {
diff = difficulty;
//easy = 0; medium = 1; hard = 2
try (Scanner scanFile  = new Scanner(new File(fileName))) {
while(scanFile.hasNext()) {
String line = scanFile.nextLine();
Scanner scanLine = new Scanner(line);
String tempWord = scanLine.next();
words.add(tempWord);
}
}
catch (FileNotFoundException e) {
System.out.println("File not foundn" + e);
}
}

最后，我建议使用调试器来了解你在每个刽子手身上放了什么。