Java List合并去重

时间:2023-03-09 16:46:49
Java List合并去重

List A和B

A.removeAll(B);
A.addAll(B);

例如有如下实体类:


/**
* hashset是如何保持元素的唯一性呢?
* 是通过元素的hashcode和equals来表示:
* 如果hashCode值一样,则比较equals是否为true
* 如果hashCode值不一样,不用比较equals
*/
/**
* List是如何集合中元素相同的呢?
* 是通过元素的hashcode和equals来表示:
* 如果hashCode值一样,则比较equals是否为true
* 如果hashCode值不一样,不用比较equals
*/
public class UserTable {
private String linkdoodid; private String linkdoodname;   public UserTable() {
super();
}
  public UserTable(String linkdoodid,String linkdoodname){
    supert();
    this.linkdoodid=linkdoodid;
    this.linkdoodname=linkdoodname;
  } public String getLinkdoodid() {
return linkdoodid;
} public void setLinkdoodid(String linkdoodid) {
this.linkdoodid = linkdoodid == null ? null : linkdoodid.trim();
} public String getLinkdoodname() {
return linkdoodname;
} public void setLinkdoodname(String linkdoodname) {
this.linkdoodname = linkdoodname == null ? null : linkdoodname.trim();
} @Override
public boolean equals(Object obj) {
if (!(obj instanceof UserTable)) {
return false;
}
UserTable userTable = (UserTable) obj;
return this.linkdoodid.equals(userTable.linkdoodid);
} @Override
public int hashCode() {
return linkdoodid.hashCode();
}
}

测试:

public class HashSetTest {
public static void main(String[] args) {
    //List
    List<UserTable> listA=new ArrayList<UserTable>();
    listA.add(new UserTable("A1001","LJ"));
    listB.add(new UserTable("B1002","MH"));     List<UserTable> listB=new ArrayList<UserTable>();
    listB.add(new UserTable("B1002","SM"));
    listB.add(new UserTable("C1001","TM"));
    
    listA.removeAll(listB);//由于UserTable的hashCode和equal 都是以linkdoodid 来判断,所以“B1002”算重复元素    
    listA.addAll(listB);     //HashSet
   HashSet<UserTable> hs = new HashSet<UserTable>();
   hs.add(new UserTable("a1", 20));
   hs.add(new UserTable("a2", 30));
   hs.add(new UserTable("a3", 40));
   hs.add(new UserTable("a3", 40));
   Iterator<Person> iterator = hs.iterator();
   while(iterator.hasNext()){
   Person p = iterator.next();
   System.out.println(p.getName()+" "+p.getAge());
  }
}
}