java - 根据出现次数升序排列数组

Question

如何在 java 中根据该值的出现次数按升序排列数组中的元素。

这是我试过的:

int a[]={0,0,0,1,3,3,2,1,3,5,6,0};
int b=a.length;
for(int i=0;i<b;i++) {
    for(int j=0;j<i;j++) {
        int temp;
        if( a[j]>a[i]) {
            temp=a[i];
            a[i]=a[j];
            a[j]=temp;
        }
    }
}

for(int r=0;r<a.length;r++) {
    System.out.println(a[r]);
}

Answer 1

这是使用 TreeMap 进行此操作的有效方法。

import java.util.Comparator;
import java.util.HashMap;
import java.util.Map;
import java.util.TreeMap;

public class FrequencySort {
    public static void main(String[] args) {
        int[] ar = new int[] {5,2,8,8,5,5,8,1,9,0,1,1,0,1};

        Map<Integer,Integer> numbers = new HashMap<>();

        for(int number : ar) {
            if(numbers.containsKey(number)) {
                Integer  count = numbers.get(number);
                numbers.put(number, ++count);
            } else {
                numbers.put(number,1);
            }
        }

        final class FrequencyComparator implements Comparator<Integer> {
            Map<Integer,Integer> refMap;
            public FrequencyComparator(Map<Integer,Integer> base) {
                this.refMap = base;
            }

            @Override
            public int compare(Integer k1, Integer k2) {
                Integer val1 = refMap.get(k1);
                Integer val2 = refMap.get(k2);

                int num = val1.compareTo(val2)  ;
                // if frequencies are same then compare number itself
                return  num == 0 ? k1.compareTo(k2)   : num;
            }
        }

        FrequencyComparator comp = new FrequencyComparator(numbers);
        TreeMap<Integer,Integer> sortedMap = new TreeMap<Integer,Integer>(comp);
        sortedMap.putAll(numbers);
        for(Integer i : sortedMap.keySet()) {
            int frequencey = sortedMap.get(i);
            for(int count  = 1 ; count <= frequencey ; count++) {
                System.out.print(i + " " );
            }
        }
    }
}