java 实现多个list 合并成一个去掉重复的案例

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 extends Integer>)

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()

WaZWMkILHj .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));

}

/**

* output:map:={1=SkillUpgrade [skillId=null, skillName=null, lv=1, persNum=null, selectNum=1100, appearNum=null], 2=SkillUpgrade http://[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]]

*/

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]]

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 extends Integer>)

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()

WaZWMkILHj .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));

}

/**

* output:map:={1=SkillUpgrade [skillId=null, skillName=null, lv=1, persNum=null, selectNum=1100, appearNum=null], 2=SkillUpgrade http://[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]]

*/

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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