计算数组中数字的出现次数 [英] counting occurrence of numbers in array
问题描述
我已经有一段时间了.我试过调试,但找不到解决方案.我正在尝试计算数字的出现次数.所以我的问题是当我打印输出时它是
I am stack for a while . I tried debugging but I couldn't figure out the solution. I am trying to count the occurrence of numbers. So my problem is that when I print the output it is
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
所以如果数字出现超过 1 次,它应该只说一次,而不是出现次数多.干杯这是代码
so if the number has more than 1 occurrence it should say it only once not as many times as there is occurrences. Cheers Here is the code
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");
}
}
}
推荐答案
另一个选项是 guava 的 Multiset classes,它将为您跟踪计数:
Another option is guava's Multiset classes, which will track the count for you:
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));
这里,Multiset、HashMultiset 和 Ints 都是番石榴类.
Here, Multiset, HashMultiset and Ints are all guava classes.
请注意,Multiset 通过使用 Map 和计数器来跟踪计数器几乎可以完成上面提到的那些.它只是从您那里抽象出来,使您的代码更简单.
Note that Multiset does pretty much what someone mentioned above by using a Map and counter to track the counters. It's just abstracted away from you to make your code simpler.
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