生成随机数组String []，而无需在Java中重复（android）

Question

我正在创建android测验应用程序。 我已将其问题随机化，但有时会重复生成的问题。

显示所有问题后，如何防止其重复并终止其方法？

这是整个工作代码：

public class MainActivity extends Activity {

private int currentQuestion;
private String [] questions;
private String [] answers;
private Button answerButton;
private Button questionButton;
private TextView questionView;
private TextView answerView;
private EditText answerText; 

Random random = new Random();

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);
    init();
}

public void init() {
 questions = new String[]{"Q1?","Q2","Q3","Q4","Q5","Q6","Q7","Q8","Q9","Q10"};
 currentQuestion = random.nextInt(questions.length);
 answers = new String[]{"A1","A2","A3","A4","A5","A6","A7","A8","A9","A10"};
 answerButton = (Button)findViewById(R.id.AnswerButton);
 questionButton = (Button)findViewById(R.id.QuestionButton);
 questionView = (TextView) findViewById(R.id.QuestionTextView);
 answerView = (TextView) findViewById(R.id.AnswerTextView);
 answerText = (EditText) findViewById(R.id.AnswerText);

 answerButton.setOnClickListener(new OnClickListener(){
 @Override
 public void onClick(View v) {checkAnswer();
}});

 questionButton.setOnClickListener(new OnClickListener(){
     @Override
 public void onClick(View v) {
 showQuestion();

     }});
}

public int showQuestion(){
currentQuestion++;
if(currentQuestion == questions.length);
 currentQuestion = 0;
currentQuestion = (random.nextInt(questions.length));

questionView.setText(questions[currentQuestion]);
 answerView.setText("");
 answerText.setText("");
    return currentQuestion;
    }

public void checkAnswer() {
String answer = answerText.getText().toString();
 if(isCorrect(answer))
 answerView.setText("You're right!");
 else answerView.setText("Sorry, the correct answer is "+answers[currentQuestion]);
}
}

Answer 1

1. 给每个问题一个数字（int），并假设您有10个问题。
2. 产生1到10之间的随机数。
3. 创建一个集合来跟踪您的问题（最好列出。）
4. 每次生成随机数时，请检查集合中是否已包含该数字。 如果是，则生成另一个随机数，否则将其添加到集合中。

Answer 2

使用Fisher-Yates随机播放可在使用之前随机播放questions和answers 。

 static void fyShuffle(String[] questions, String[] answers)
 {
     Random rnd = new Random();
     for (int i = questions.length - 1; i > 0; --i){
         int index = rnd.nextInt(i + 1);
         // Swap questions
         String s = questions[index];
         questions[index] = questions[i];
         questions[i] = s;
         // Swap answers
         s = answers[index];
         answers[index] = answers[i];
         answers[i] = s;
      }
  }

参见http://en.wikipedia.org/wiki/Fisher-Yates_shuffle

它具有理想的属性，即在初始化之后（即O（N）），数组访问时间是恒定的。 拒绝重绘元素的算法会减慢速度，因此无法很好地扩展。

Answer 3

例如，您可以使用一个不包含重复项的集合（ HashSet ），并使用它代替String[] 。 如果您想要String []数组，则可以从HashSet的内容轻松创建它