No51-【原】-用实例理解HashSet

文章目录

1. 1. 前言
2. 2. 一个奇怪的问题
3. 3. 感谢

前言

众所周知,HashSet中不含有重复的元素.本文用几个例子来加深对不重复的理解.

一个奇怪的问题

先来看一段代码,代码如下:

package com.yongf.algorithm;

import java.util.HashSet;

/**
 * Created by yongf on 16-4-28.
 */
public class UnderstandingHashSet {

    public static void main(String[] args) {
        HashSet<Person> hs = new HashSet<>();
        Person person = new Person("Scott Wang", 20);
        hs.add(person);

        person.setName("王用");
        person.setAge(20);

        hs.add(person);

        System.out.println("hs = " + hs);
    }
}

class Person {
    private String name;
    private int age;

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

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

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    public int getAge() {
        return age;
    }

    public void setAge(int age) {
        this.age = age;
    }
}

运行结果为:

hs = [Person{name='王用', age=21}]

看到这里,你可能就疑问了,唉,怎么只有一个元素,不是add了两个元素么?一个是(“Scott Wang”, 20),一个是(“王用”, 20),这输出不对吧.

其实你这么分析,好像挺对,不过你忽视了一个东西,判断是否重复基于hashCode(),而在上述代码中,Person类根本就没有重写hashCode()方法,所以,HashSet在比较元素的时候根本就没有考虑name和age属性,因为继承的是Object类的hashCode()方法,所以只会比较对象,只要是同一个对象,在不重写hashCode()方法的情况下,都被认为是同一个对象.

好吧,那我们重写hashCode()方法,代码如下:

package com.yongf.algorithm;

import java.util.HashSet;

/**
 * Created by yongf on 16-4-28.
 */
public class UnderstandingHashSet {

    public static void main(String[] args) {
        HashSet<Person> hs = new HashSet<>();
        Person person = new Person("Scott Wang", 20);
        hs.add(person);

        person.setName("王用");
        person.setAge(20);

        hs.add(person);

        System.out.println("hs = " + hs);
    }
}

class Person {
    private String name;
    private 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 int hashCode() {
        return name.hashCode() + age * 100;
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    public int getAge() {
        return age;
    }

    public void setAge(int age) {
        this.age = age;
    }
}

现在hashCode()方法与name,age属性挂钩上了,在此运行,结果如下:

hs = [Person{name='王用', age=20}, Person{name='王用', age=20}]

不过,这里的结果还是两个一样的,我的想法如下图所示:

不过,如果创建一个新的person1 = new Person(“Scott Wang”, 20),然后hs.add(person1),结果如下:

hs = [Person{name='Scott Wang', age=20}, Person{name='Scott Wang', age=20}]

这显然也是不符合我们预期的,如果name和age都一样,我们预期两者是不同的,为此,我们需要重写equals方法,因为HashSet的判重涉及到了hashCode(), equals()方法, 于是,经过我们的两次修正,最后的代码如下:

package com.yongf.algorithm;

import java.util.HashSet;

/**
 * Created by yongf on 16-4-28.
 */
public class UnderstandingHashSet {

    public static void main(String[] args) {
        HashSet<Person> hs = new HashSet<>();
        Person person = new Person("Scott Wang", 20);
        hs.add(person);

        Person person1 = new Person("Scott Wang", 20);

        hs.add(person1);

        hs.add(new Person("老码", 20));
        hs.add(new Person("王用", 20));

        System.out.println("hs = " + hs);
    }
}

class Person {
    private String name;
    private 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 int hashCode() {
        return name.hashCode() + age * 100;
    }

    @Override
    public boolean equals(Object obj) {
        if (obj instanceof Person) {
            Person person = (Person) obj;
            return person.name.equals(name) && (person.age == age);
        } else {
            return false;
        }
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    public int getAge() {
        return age;
    }

    public void setAge(int age) {
        this.age = age;
    }
}

最后的运行结果为:

hs = [Person{name='王用', age=20}, Person{name='Scott Wang', age=20}, Person{name='老码', age=20}]

OK,这个结果在预料之中.那么,我们下次再见 =_=

感谢

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