TreeMap.get() return Null 偶数键存在 [英] TreeMap.get() return Null even key exists

问题描述

我试图从 TreeMap 获取，但即使键存在它也返回 null.HashCode 和 eqauls 仅基于单词.可比性基于频率.

I am trying to get from TreeMap but it return null even the key exist. HashCode and eqauls is based on word only. Comparable is based on freqency.

    public static void main(){
        TreeMap<Word,Integer> test = new TreeMap<>();
        test.put(new Word("pqr",12),1);
        test.put(new Word("abc",2),1);

        Integer prq = test.get(new Word("pqr",1));
        System.out.println(prq);
        prq = test.get(new Word("pqr",12));
        System.out.println(prq);
    }


    public class Word implements Comparable<Word>{
        String word;
        Integer freq;

        public Word(String word, Integer freq) {
            this.word = word;
            this.freq = freq;
        }

        @Override
        public boolean equals(Object o) {
            if (this == o) return true;
            if (!(o instanceof Word)) return false;

            Word word1 = (Word) o;

            return word.equals(word1.word);
        }

        @Override
        public int hashCode() {
            return word.hashCode();
        }


        @Override
        public int compareTo(Word o) {
            return this.freq.compareTo(o.freq);
        }
    }

输出就像空值1

推荐答案

在您的 compareTo 方法中，它正在比较频率.因此，如果频率相同，它将相等.

In your compareTo method it is comparing the frequency. So if the frequency is same it will be equal.

比较你可以使用的词

return this.word.compareTo(o.word);

或者比较你可以使用的词和频率

or to compare both word and frequency you can use

return this.word.compareTo((o.word)) *  this.freq.compareTo(o.freq);

编辑

现在您需要使用频率进行排序，因此您可以使用 Comparator 而不是使用可比较.使用上述比较器创建 Map.并使用您以前的比较器进行排序.

Now as you need to sort using the frequency so instead of using comparable you can use Comparator. Use the above comparators to create the Map. And use your previous comprator to sort.

在创建时

        TreeMap<Word, Integer> test = new TreeMap<Word, Integer>(
            new Comparator<Word>() {
                public int compare(Word word, Word o) {
                    return word.word.compareTo((o.word));
                }
            });

并且在排序时

    Collections.sort(new LinkedList(test.keySet()), new Comparator<Word>() {
        public int compare(Word word, Word o) {
            return word.freq.compareTo((o.freq));
        }
    });

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