java8两个List集合取交集、并集、差集、去重并集

import java.util.ArrayList;
import java.util.List;
import static java.util.stream.Collectors.toList;
 
public class Test {
 
    public static void main(String[] args) {
        List<String> list1 = new ArrayList<String>();
        list1.add("1");
		list1.add("2");
		list1.add("3");
		list1.add("5");
		list1.add("6");
 
        List<String> list2 = new ArrayList<String>();
        list2.add("2");
		list2.add("3");
		list2.add("7");
		list2.add("8");
 
        // 交集
        List<String> intersection = list1.stream().filter(item -> list2.contains(item)).collect(toList());
        System.out.println("---交集 intersection---");
        intersection.parallelStream().forEach(System.out :: println);
 
        // 差集 (list1 - list2)
        List<String> reduce1 = list1.stream().filter(item -> !list2.contains(item)).collect(toList());
        System.out.println("---差集 reduce1 (list1 - list2)---");
        reduce1.parallelStream().forEach(System.out :: println);
 
        // 差集 (list2 - list1)
        List<String> reduce2 = list2.stream().filter(item -> !list1.contains(item)).collect(toList());
        System.out.println("---差集 reduce2 (list2 - list1)---");
        reduce2.parallelStream().forEach(System.out :: println);
 
        // 并集
        List<String> listAll = list1.parallelStream().collect(toList());
        List<String> listAll2 = list2.parallelStream().collect(toList());
        listAll.addAll(listAll2);
        System.out.println("---并集 listAll---");
        listAll.parallelStream().forEachOrdered(System.out :: println);
 
        // 去重并集
        List<String> listAllDistinct = listAll.stream().distinct().collect(toList());
        System.out.println("---得到去重并集 listAllDistinct---");
        listAllDistinct.parallelStream().forEachOrdered(System.out :: println);
 
        System.out.println("---原来的List1---");
        list1.parallelStream().forEachOrdered(System.out :: println);
        System.out.println("---原来的List2---");
        list2.parallelStream().forEachOrdered(System.out :: println);
 
    }
}