计算数组中数字的出现

Question

我堆了一会儿。我尝试调试，但找不到解决方案。我正在尝试计算数字的出现。所以我的问题是当我打印输出时

3 occurs 1 times
1 occurs 1 times
0 occurs 1 times
2 occurs 1 times
1 occurs 2 times
3 occurs 2 times
2 occurs 2 times
0 occurs 2 times
10 occurs 1 times
4 occurs 1 times

代替

1 occurs 2 times
0 occurs 2 times
2 occurs 2 times
3 occurs 2 time
10 occurs 1 times
4 occurs 1 times

因此，如果数字出现的次数超过1，则应该只说一次，而不是出现的次数。欢呼这是代码

import java.util.*;

public class CountingOccuranceOfNumbers
{

    public static void main(String[] args) 
    {
        countNumbers();
    }

    public static void countNumbers()
    {
        Scanner input = new Scanner(System.in);
        Random generator = new Random();
        int[] list = new int[11];
        int[] counts = new int[150];
        int counter = 0;
        int number = 1;


        while(counter <= 10)
        {
                number = generator.nextInt(11);
                list[counter] = number;
                counter++;
        }   
        for(int i=0; i<list.length - 1; i++)
        {
            counts[list[i]]++;
//          System.out.print(list[i] + " ");

            System.out.println(list[i] +" occurs " +  counts[list[i]] + " times");
        }

    }

}

Answer 1

另一个选项是番石榴的Multiset类，它将为您跟踪计数：

int values[] = ...;
Multiset<Integer> ms = HashMultiset.create();
ms.addAll(Ints.asList(list));

int count0 = ms.count(Integer.valueOf(0));
int count1 = ms.count(Integer.valueOf(1));

在这里，Multiset，HashMultiset和Ints都是番石榴类。

请注意，Multiset通过使用Map和counter来跟踪计数器，几乎可以完成上面提到的某些事情。它只是从您身上抽象出来，以使您的代码更简单。