快速从两个集合中查重

如果给你两个集合，需要挑选出重复的元素，怎么解决？

两个for循环？依次遍历？那样太慢了。

package com.cw.wizbank;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

/**
 * @Description:
 * @Author :小书包
 * @CreateDate :2018-11-21 10:10
 */
public class DuplicatedList {

    public static void main(String[] args) {
        List list1 = new ArrayList<>(Arrays.asList(new Person("jack",12)
                ,new Person("smith",14)
                ,new Person("tom",12)
                ,new Person("alex",13)
                ,new Person("red",12)
                ,new Person("green",16)
        ));

        List list2 = new ArrayList<>(Arrays.asList(new Person("jack",12)
                ,new Person("smith",14)
                ,new Person("tom1",12)
                ,new Person("alex1",13)
                ,new Person("red",12)
                ,new Person("green",16)
        ));

        boolean boo = list1.removeAll(list2);
        System.out.println(boo);
        System.out.println(list1);
        System.out.println(list2);
    }

}
class Person {
    private String name;
    private int age;

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

    @Override
    public boolean equals(Object o) {
        if (this == o) return true;
        if (o == null || getClass() != o.getClass()) return false;

        Person person = (Person) o;

        if (age != person.age) return false;
        return name != null ? name.equals(person.name) : person.name == null;

    }

    @Override
    public int hashCode() {
        int result = name != null ? name.hashCode() : 0;
        result = 31 * result + age;
        return result;
    }

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

一句代码就解决查重问题。

boolean boo = list1.removeAll(list2);

当然在这个里面，因为是对对象进行判断的，因此需要对象实现equals方法和hashcode方法

true
[Person{name='tom', age=12}, Person{name='alex', age=13}]
[Person{name='jack', age=12}, Person{name='smith', age=14}, Person{name='tom1', age=12}, Person{name='alex1', age=13}, Person{name='red', age=12}, Person{name='green', age=16}]

查询之后的结果如上