java 实现多个list 合并成一个去掉重复的案例

我就废话不多说了，大家还是直接看代码吧~

 public static void main(String[] args){
    List list1 = new ArrayList();
    list1.add(1);
    list1.add(2);
    list1.add(3);
    list1.add(4);
    List list2 = new ArrayList();
    list2.add(1);
    list2.add(4);
    list2.add(7);
    list2.add(10);
    List listAll = new ArrayList();
    listAll.addAll(list1);
    listAll.addAll(list2);
    listAll = new ArrayList(new linkedHashSet<>(listAll));
    System.out.println(listAll);
  }

输出：

[1, 2, 3, 4, 7, 10]

代码要典：

1、合并 使用java.util.List.addAll(Collection)

2、去重,借助linkedHashSet

补充知识：java8 lambda小试牛刀，利用Stream把list转map,并将两个list的数据对象合并起来

我就废话不多说了，大家还是直接看代码吧~

public static void main(String[] args) {
		// 集合1
		List lists = new ArrayList<>();
		SkillUpgrade s = new SkillUpgrade();
		s.setLv(1);
		s.setAppearNum(100);
		lists.add(s);
		SkillUpgrade s2 = new SkillUpgrade();
		s2.setLv(2);
		s2.setAppearNum(200);
		lists.add(s2);
		// 集合1
		List listx = new ArrayList<>();
		SkillUpgrade x = new SkillUpgrade();
		x.setLv(1);
		x.setSelectNum(1100);
		listx.add(x);
		SkillUpgrade x2 = new SkillUpgrade();
		x2.setLv(2);
		x2.setSelectNum(1200);
		listx.add(x2);
		// 把list转map,{k=lv,vaule=并为自身} . SkillUpgrade->SkillUpgrade或Function.identity()
		Map map = listx.stream()
				.collect(Collectors.toMap(SkillUpgrade::getLv, SkillUpgrade -> SkillUpgrade));
		System.out.println("map:="+map);
		// 合并
		lists.forEach(n -> {
			// 如果等级一致
			if (map.containsKey(n.getLv())) {
				SkillUpgrade obj = map.get(n.getLv());
				// 把数量复制过去
				n.setSelectNum(obj.getSelectNum());
			}
		});
		System.out.println("lists:="+lists);
		// 重复问题
		Map keyRedo = listx.stream()
				.collect(Collectors.toMap(SkillUpgrade::getLv, Function.identity(), (key1, key2) -> key2));
		// 方式二：指定实例的map
		Map linkedHashMap = listx.stream().collect(Collectors.toMap(SkillUpgrade::getLv,
				SkillUpgrade -> SkillUpgrade, (key1, key2) -> key2, linkedHashMap::new));
	}
	
	

输出结果：

map:={1=SkillUpgrade [skillId=null, skillName=null, lv=1, persNum=null, selectNum=1100, appearNum=null], 2=SkillUpgrade [skillId=null, skillName=null, lv=2, persNum=null, selectNum=1200, appearNum=null]}

lists:=[SkillUpgrade [skillId=null, skillName=null, lv=1, persNum=null, selectNum=1100, appearNum=100], SkillUpgrade [skillId=null, skillName=null, lv=2, persNum=null, selectNum=1200, appearNum=200]]

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