Question

到目前为止我的代码：

 public void lottoGame() {        
      HashSet<Integer> dups = new HashSet<Integer>();  

      for(int howMany = 1; howMany <= 6; howMany++) {
           int lottoNumber = lottoNum.nextInt(49) + 1;

           while(dups.contains(lottoNumber)) {
                lottoNumber = lottoNum.nextInt(49) + 1;
           } 

           dups.add(lottoNumber);
           randomLotto[howMany] = lottoNumber;

           System.out.println(lottoNumber);

    }
    for(int counter : dups) {
        numberCount[counter]++; //to store the counting of random numbers
    } 

    for(int counting = 1; counting < numberCount.length; counting++) {
        System.out.println(counting + " occurs " + numberCount[counting] + " times");
    }
}

基本上，我所做的是创建一个hashset并输入6个随机数。在我的最后2个for循环中，我试图计算我绘制哪个数字并打印出来的频率。问题是，无论出于何种原因，它在打印时从第3个字段开始。有谁知道为什么？

Answer 1

以下是您提供的示例。我的输出是正确的。验证您可能未向我们展示的设置或其他代码。

请注意：Is there a limit on the maximum of number of lines which can be printed to console by BlueJ?

class Test
{
    public void lottoGame()
    {
        int[] randomLotto = new int[50];
        int[] numberCount = new int[50];
        Random lottoNum = new Random();
        HashSet<Integer> dups = new HashSet<Integer>();

        for (int howMany = 1; howMany <= 6; howMany++)
        {
            int lottoNumber = lottoNum.nextInt(49) + 1;

            while (dups.contains(lottoNumber))
            {
                lottoNumber = lottoNum.nextInt(49) + 1;
            }

            dups.add(lottoNumber);
            randomLotto[howMany] = lottoNumber;

            System.out.println(lottoNumber);

        }
        for (int counter : dups)
        {
            numberCount[counter]++; //to store the counting of random numbers
        }
        for (int counting = 1; counting < numberCount.length; counting++)
        {
            System.out.println(counting + " occurs " + numberCount[counting] + " times");
        }
    }

    public static void main(String[] args)
    {
        new Test().lottoGame();
    }
}

输出：

43
27
38
30
14
8
1 occurs 0 times
2 occurs 0 times
3 occurs 0 times
4 occurs 0 times
5 occurs 0 times
6 occurs 0 times
7 occurs 0 times
8 occurs 1 times
9 occurs 0 times
10 occurs 0 times
11 occurs 0 times
12 occurs 0 times
13 occurs 0 times
14 occurs 1 times
15 occurs 0 times
16 occurs 0 times
17 occurs 0 times
18 occurs 0 times
19 occurs 0 times
20 occurs 0 times
21 occurs 0 times
22 occurs 0 times
23 occurs 0 times
24 occurs 0 times
25 occurs 0 times
26 occurs 0 times
27 occurs 1 times
28 occurs 0 times
29 occurs 0 times
30 occurs 1 times
31 occurs 0 times
32 occurs 0 times
33 occurs 0 times
34 occurs 0 times
35 occurs 0 times
36 occurs 0 times
37 occurs 0 times
38 occurs 1 times
39 occurs 0 times
40 occurs 0 times
41 occurs 0 times
42 occurs 0 times
43 occurs 1 times
44 occurs 0 times
45 occurs 0 times
46 occurs 0 times
47 occurs 0 times
48 occurs 0 times
49 occurs 0 times

Answer 2

我建议使用map，这个例子看起来更简单

  public static void lotoGame() {
    Random rand = new Random();
    Map<Integer, Integer> loto = new HashMap<>();
    while (loto.keySet().size() < 6) {
        int lottoNumber = rand.nextInt(49) + 1;
        loto.compute(lottoNumber, (key, value) -> value == null ? 1 : ++value);
    }

    loto.forEach((key,value)->{
        System.out.println(key + " occurs " + value + " times");
    });
}

计算拾取Hashset中的数字的频率

2 个答案: