Arrays.sort引发编译时错误“派生自Comparator ..的匿名类." [英] Arrays.sort throws compile-time error 'Anonymous class derived from Comparator..'

问题描述

错误:

必须声明从Comparator派生的匿名类抽象或实现抽象方法compare(T，T).

Anonymous class derived from Comparator must either be declared Abstract or implement abstract method compare(T,T).

我想根据地图中的频率对数组进行排序.

I want to sort the Array based on frequency in the map.

HashMap<Integer,Integer> freq = new HashMap<>();
        for(int a:arr)
        {
            freq.putIfAbsent(a,freq.getOrDefault(a,0)+1);
        }
Arrays.sort(arr,new Comparator<>(){
            public int compare(int a,int b)
            {
                if(freq.get(b)!=freq.get(a))
                    return freq.get(b)-freq.get(a);
                else
                    return a-b;
            }
        });

我在这里想念什么?

推荐答案

假设您有 Integer [] 之类的

Integer[] arr = { 10, 2, 30, 4 };

您可以按照以下步骤进行操作:

you can do it as follows:

Arrays.sort(arr, new Comparator<Integer>() {
    @Override
    public int compare(Integer a, Integer b) {
        if (freq.get(b) != freq.get(a))
            return Integer.compare(freq.get(b), freq.get(a));
        else
            return a.compareTo(b);
    }
});

如果您有 int [] 这样的

int[] arr = { 10, 2, 30, 4 };

将其转换为 Integer [] as

Integer[] arr1 = Arrays.stream(arr).boxed().toArray(Integer[]::new);

，然后将 arr1 排序为

Arrays.sort(arr1, new Comparator<Integer>() {
    @Override
    public int compare(Integer a, Integer b) {
        if (freq.get(b) != freq.get(a))
            return Integer.compare(freq.get(b), freq.get(a));
        else
            return a.compareTo(b);
    }
});

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