简单的理解集合中的HashSet和HashTree几个重写方法

Java中的set是无序的,但是是不可重复的
HashSet底层是哈希表,通过调用hashcode和equals方法实现去重
当我们HashSet里面存的是字符串时,就能默认去重了,因为String已经重写了hashcode和euqals方法

    public static void main(String[] args) {
        HashSet<String> set = new HashSet();
        set.add("java");
        set.add("c");
        set.add("php");
        set.add("bigdata");
        set.add("java");
        //运行结果,给去重了,而且是无序的
        System.out.println(set);//[java, c, bigdata, php]
    }
}

但是当我们有类的时候,比如Person,Cat,Dog,我们自己写的类,但是我们想按照自己制定的规则去重,就比如Person的姓名和年龄,因为Person类是我们自己建的,如果我们没有重写方法,就会去找Object的hashcode方法,这样new Person()的hashcode就会不同,这样每个都是一个新的,都会输出,即使年龄和姓名一样

public class Demo1 {
    
    
    public static void main(String[] args) {
    
    
        HashSet<Person> set1 = new HashSet<>();
        set1.add(new Person("aing",50));
        set1.add(new Person("bing",10));
        set1.add(new Person("ding",20));
        set1.add(new Person("ding",20));
        //运行结果是即使名字年龄一样,也会输出两个,我们如果想要按照自己的规则去重,这样我们一定要重写hashcode 和euqals方法
        System.out.println(set1);//[Person{name='bing', age=10}, Person{name='ding', age=20}, Person{name='aing', age=50}, Person{name='ding', age=20}]
    }
}
class Person{
    
    
    String name;
    int age;

    public Person(String name, int age) {
    
    
        this.name = name;
        this.age = age;
    }

    @Override
    public String toString() {
    
    
        return "Person{" +
                "name='" + name + '\'' +
                ", age=" + age +
                '}';
    }
}

我们重写hashcode方法,当hashcode不一样时,就不会比较equals,直接就是不一样,如果hashcode一样,再比较equals方法

public class Demo1 {
    
    
    public static void main(String[] args) {
    
    
        HashSet<Person> set1 = new HashSet<>();
        set1.add(new Person("aing",50));
        set1.add(new Person("bing",10));
        set1.add(new Person("ding",20));
        set1.add(new Person("ding",20));
        //根据年龄和姓名比较的
        System.out.println(set1);//[Person{name='ding', age=20}, Person{name='aing', age=50}, Person{name='bing', age=10}]
    }
}
class Person{
    
    
    String name;
    int age;
    public Person(String name, int age) {
    
    
        this.name = name;
        this.age = age;
    }
    @Override
    public String toString() {
    
    
        return "Person{" +
                "name='" + name + '\'' +
                ", age=" + age +
                '}';
    }
    @Override
    public boolean equals(Object o) {
    
    
        if (this == o) return true;
        if (o == null || getClass() != o.getClass()) return false;
        Person person = (Person) o;
        return age == person.age &&
                Objects.equals(name, person.name);
    }
    //和属性关联,根据属性的值比较,之所以让age*100;是因为怕有可能出现这个name.hashCode()+age 和 下一个name.hashCode()+age加起来的值恰好相等,所以age*100,可以避免这种情况的发生
    @Override
    public int hashCode() {
    
    
        //return Objects.hash(name, age);
        return name.hashCode()+age*100;
    }
}

TreeSet底层是二叉树,而且TreeSet还是有序的,在String中不仅重写了hashcode,还要调用元素的compareTo方法, String类已经实现了Comparable接口,并重写 了compareTo方法,但是如果我们自己写类的话,比如Person来说,如果想要按照自己的规则比,就要重写hashcode方法和实现Comparable接口

public class Demo2 {
    
    
    public static void main(String[] args) {
    
    
        TreeSet<String> set = new TreeSet();
        set.add("java");
        set.add("c");
        set.add("php");
        set.add("bigdata");
        set.add("java");
        System.out.println(set);//[bigdata, c, java, php]
    }
}

但是要是自己的类
会报错cannot be cast to java.lang.Comparable,因为Person会去找compareTo的方法,但是Person类没有实现它,我们要去实现Comparable

public class Demo2 {
    
    
    public static void main(String[] args) {
    
    
        TreeSet<Person1> set1 = new TreeSet<>();
        set1.add(new Person1("aing",50));
        set1.add(new Person1("bing",10));
        set1.add(new Person1("ding",20));
        set1.add(new Person1("ding",20));
        System.out.println(set1);
    }
}
class Person1{
    
    
    String name;
    int age;

    public Person1(String name, int age) {
    
    
        this.name = name;
        this.age = age;
    }

    @Override
    public String toString() {
    
    
        return "Person1{" +
                "name='" + name + '\'' +
                ", age=" + age +
                '}';
    }
}

这样我们就可以实现有序了,根据我们的属性

public class Demo2 {
    
    
    public static void main(String[] args) {
    
    
        TreeSet<Person1> set1 = new TreeSet<>();
        set1.add(new Person1("aing",50));
        set1.add(new Person1("bing",10));
        set1.add(new Person1("ding",20));
        set1.add(new Person1("ding",20));
        System.out.println(set1);
    }
}
class Person1 implements Comparable{
    
    
    String name;
    int age;

    public Person1(String name, int age) {
    
    
        this.name = name;
        this.age = age;
    }

    @Override
    public boolean equals(Object o) {
    
    
        if (this == o) return true;
        if (o == null || getClass() != o.getClass()) return false;
        Person1 person1 = (Person1) o;
        return age == person1.age &&
                Objects.equals(name, person1.name);
    }

    @Override
    public int hashCode() {
    
    
        return name.hashCode()+age*100;
    }

    @Override
    public String toString() {
    
    
        return "Person1{" +
                "name='" + name + '\'' +
                ", age=" + age +
                '}';
    }
    //根据类的属性进行排序
    @Override
    public int compareTo(Object o) {
    
    
    Person1 person = (Person1)o;
    int num = name.compareTo(person.name);
    return num==0?age-person.age:num;
}
}

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转载自blog.csdn.net/m0_51327764/article/details/109018469
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