2011-10-08 164 views
0

如何让它在4行中的任意一行上随机显示答案,而不会有任何重复?无法使此代码正常工作

我当前的代码:这是负责对4条线路(随机)

answer1.setText(answers_list[Type][randomValue+1 > 3 ? (randomValue+0)-4 : randomValue+0]); 
    answer2.setText(answers_list[Type][randomValue+2 > 3 ? (randomValue+1)-3 : randomValue+1]); 
    answer3.setText(answers_list[Type][randomValue+3 > 3 ? (randomValue+2)-2 : randomValue+2]); 
    answer4.setText(answers_list[Type][randomValue+0 > 3 ? (randomValue+3)-4 : randomValue+3]); 

目前一个采摘&添加问题

TextView question; 
private int qType = -1; 
private int asked = 0; 
private void QBegin() { 
    /* 
    * Gets a random question 
    */ 
    question = (TextView) findViewById(R.id.question); 
    String[] types = { "Q1", "Q2", "Q3", "Q4", "Q5"}; 
    Random random = new Random(); 
    int qType = random.nextInt(types.length); 
    question.setText(types[qType]); 
    asked++; 
    // StringList.add(types[qType]); 
    getAnswers(qType); 
/* if(StringList.contains(types[qType]) && asked >= types.length+1){ 
     asked = 0; 
     answerCounter.setText("THE END"); 
    } else if (StringList.contains(types[qType]) && asked < types.length+1){ 
     QBegin(); 
    } */ 
} 
    public static int random(int range) { 
    return (int)(java.lang.Math.random() * (range+1)); 
} 

public void shuffle(String input){ 
    /* 
    * Unused shuffle method 
    */ 
    List<Character> characters = new ArrayList<Character>(); 
    for(char c:input.toCharArray()){ 
     characters.add(c); 
    } 
    StringBuilder output = new StringBuilder(input.length()); 
    while(characters.size()!=0){ 
     int randPicker = (int)(Math.random()*characters.size()); 
     output.append(characters.remove(randPicker)); 
    } 
    System.out.println(output.toString()); 
} 
private void getAnswers(int Type) { 
    /* 
    * Getting answers here 
    */ 
    int randomValue = random(4); 
    try { 
    String answers_list[][] = { 
      {"Answer 1-1", "Answer 2-1", "Answer 3-1", "Answer 4-1"}, 
      {"Answer 1-2", "Answer 2-2", "Answer 3-2", "Answer 4-2"}, 
      {"Answer 1-3", "Answer 2-3", "Answer 3-3", "Answer 4-3"}, 
      {"Answer 1-4", "Answer 2-4", "Answer 3-4", "Answer 4-4"}, 
      {"Answer 1-5", "Answer 2-5", "Answer 3-5", "Answer 4-5"}} ; 
    answer1.setText(answers_list[Type][randomValue+1 > 3 ? (randomValue+0)-4 : randomValue+0]); 
    answer2.setText(answers_list[Type][randomValue+2 > 3 ? (randomValue+1)-3 : randomValue+1]); 
    answer3.setText(answers_list[Type][randomValue+3 > 3 ? (randomValue+2)-2 : randomValue+2]); 
    answer4.setText(answers_list[Type][randomValue+0 > 3 ? (randomValue+3)-4 : randomValue+3]); 


    /*for (int rows = 0; rows < answer&*list.length; rows++){ 
    for (int cols = 0; cols < answers_list[rows].length; cols++){ 
     } 
     }*/ 
    } catch(Exception ex){ 
    answer1.setText("Error "+ex); 
    } 



} 

代码,我与ANSWER1并具有重复answer4。请帮忙。

+0

这是...不清楚这里发生了什么,或者你如何实施这个。基本上你试图得到一个没有重复的四个随机数列表? –

+0

是的,这就是我想要做的。 – Alex

回答

1

有一种已知的标准混洗方法。

  1. 随机挑选其中一个。如有必要,将您选择的那个与第四个位置的答案进行交换。
  2. 随机选择剩余答案1至3中的一个。如有必要,将您选择的那个与第三个位置的答案进行交换。
  3. 随机选择其余答案1至2中的一个。如有必要,将您选择的那个与第二个位置的答案进行交换。

您现在有一个随机顺序包含原始四个答案的混洗列表。该算法被称为Fisher-Yates shuffle

ETA简单的示例代码:

Random rand = new Random(); 

String[] answers = { "42", 
        "Only on a Tuesday.", 
        "$4.36", 
        "Hieronymous K. Sluggenheimer III" }; 

// Shuffle answers[] 
for (int i = 3; i > 0; --i) { 
    // Pick an answer that hasn't yet been chosen. 
    int pick = rand.nextInt(i + 1); 
    if (pick != i) { 
     // Exchange answers[i] and answers[pick]. 
     String temp = answers[i]; 
     answers[i] = answers[pick]; 
     answers[pick] = temp; 
    } 
} 
+0

姆米,你能否为我修改我的代码,导致你的陈述对我有点困惑。 – Alex

+0

老实说:这对我没有太大帮助:/ – Alex

1

如果我正确理解你的目标,你想,随机混合的问题清单。 Rossum给了你一个方法来做到这一点,this是一个很好的介绍最知名的选项。

在我看来,这是最简单的方法:

  1. 创建地图的排序由键(如TreeMap <Double,String><Double,Question>
  2. 添加一个随机密钥和问题作为值
  3. 循环通过并打印结果。一个foreach循环将使用排序顺序

不要担心重复的随机数或规范化它。 Dups不太可能,如果你需要一个问题#,在循环中添加一个计数器。