为什么使用自定义Comparator到treeSet会破坏String对象的相等性？

Question

我编写了一个比较器，它计算给定字符串中的辅音来比较2个字符串，但它以某种方式使TreeSet认为2个字符串相等（可能是它们具有相同的大小）。

public class SortDiff {
public static void main(String[] s1) {

    List<String> list = Arrays.asList("century", "army", "spit", "twin", "frog", "guideline", "impulse",
            "distributor", "hallway", "appeal", "fitness", "pudding", "mild", "enter", "kitchen", "constitutional",
            "die", "condition", "morsel", "jealous", "colorful", "myth", "belly", "rugby", "valid", "shot",
            "locate", "preference", "representative", "chart", "prince", "think", "threshold", "health", "sweater",
            "volume", "poison", "lease", "toast", "diplomat", "trait", "cower", "slime", "handy", "example",
            "sanctuary", "board", "crash", "large", "attract", "censorship", "room", "license", "smoke", "roll",
            "taste", "inflate", "continuation", "establish", "fault", "gown", "dirty", "width", "qualify",
            "reference", "excitement", "vegetable", "wear", "confusion", "concept", "willpower", "snarl", "develop",
            "integration", "pie", "respectable", "fast", "limit", "shaft", "acceptance", "insert", "brown", "death",
            "effective", "ticket", "play", "highway", "lot", "tablet", "mother", "pier", "facility", "match",
            "animal", "sport", "laundry", "negligence", "white", "vat", "nuclear");
    System.out.println(list.size() + "=====");
    TreeSet<String> tree = new TreeSet<String>(new sortByConsonants());//
    int count = 0;
    for (String s : list) {
        count++;
        System.out.println(s);
        System.out.println(tree.add(s));
    }
    System.out.println("===>" + count);
    System.out.println(tree.size() + "=====");
    Iterator itr = tree.iterator();
    while (itr.hasNext()) {
        System.out.println(itr.next());
    }
}


}

这是我传递给TreeSet的比较器类。

class sortByConsonants implements Comparator<String>{

    public int compare(String a, String b) 
    { 

        return count(a,2)-count(b,2);
    } 
    public boolean equals(String a,String b) {

        return a.equals(b);

    }

    public int count(String line,int type) {
        int vowels = 0, consonants = 0, digits = 0, spaces = 0;

        line = line.toLowerCase();
        for(int i = 0; i < line.length(); ++i)
        {
            char ch = line.charAt(i);
            if(ch == 'a' || ch == 'e' || ch == 'i'
                || ch == 'o' || ch == 'u') {
                ++vowels;
            }
            else if((ch >= 'a'&& ch <= 'z')) {
                ++consonants;
            }
            else if( ch >= '0' && ch <= '9')
            {
                ++digits;
            }
            else if (ch ==' ')
            {
                ++spaces;
            }
        }

        if(type==1) {
            return vowels;
        }else {
            return consonants;
        }

    }

}

有人可以调查一下，发现我犯了什么错！

目标：我想根据辅音计数对字符串进行排序（不使用库sort（）方法）

编辑：更改为比较器逻辑

public int compare(String a, String b) 
    { 
        if(count(a,2)-count(b,2)==0) {
            return 1;
        }
        return count(a,2)-count(b,2);
    } 

还是行不通！！

编辑=对于这个问题，TreeSet不是正确的数据结构，应该使用ArrayList。

Answer 1

将您的SortByConsonants更改为以下内容。 （注意，类名应以大写字母开头）

class SortByConsonants implements Comparator<String> {

    public int compare(String a, String b) {
        if (count(a) - count(b) == 0) {
            return 1;
        }
        return count(a) - count(b);
    }

    public boolean equals(String a, String b) {
        return a.equals(b);
    }

    public int count(String line) {
        int consonants = 0;
        line = line.toLowerCase();
        for (int i = 0; i < line.length(); ++i) {
            char ch = line.charAt(i);
            if (ch != 'a' && ch != 'e' && ch != 'i' && ch != 'o' && ch != 'u') {
                consonants++;
            }
        }
        return consonants;
    }

}

此外，由于我们只对基于辅音计数的排序感兴趣，因此无需计算元音，空格，数字等的数量。此外，还不需要将另一个参数类型传递给方法count

Answer 2

你编写它的comperator将返回0（意思是相等） 如果2个字符串具有相同数量的辅音。 你定义了：

TreeSet<String> tree = new TreeSet<String>(new sortByConsonants());

所以在tree中不能存在2个具有相同辅音数量的项目（字符串）