JAVA中List數組判斷是否有重復元數

JAVA中List數組判斷是否有重復元數

1.使用Set的特性，將List轉Set,比較兩個集合的size()的大小是否相等

List<String> strList=new ArrayList<>(Arrays.asList("張三","李四","李四","王五"));
Set<String> strSet=new HashSet<>(strList);
if (strList.size() == strSet.size()) {
    System.out.println("沒有重復");
} else {
    System.out.println("重復了，老鐵");
}

2.使用jdk8---Stream的特性

List<String> strList=new ArrayList<>(Arrays.asList("張三","李四","李四","王五"));
long count = strList.stream().distinct().count();
if (strList.size() == count) {
    System.out.println("沒有重復");
} else {
    System.out.println("重復了，老鐵");
}

3.使用Object時判斷成員變量是否有重復

User類：

@Data
public class User {
    private Long id;
    private String name;
    private Integer age;

    public User(Long id, String name, Integer age) {
        this.id = id;
        this.name = name;
        this.age = age;
    }

    public User() {
    }
}

判斷User集合中name是否有重復：

List<User> users = new ArrayList<User>(){
    {
        add(new User(13131l,"張三",33));
        add(new User(21244l,"李四",23));
        add(new User(31865l,"王五",41));
        add(new User(31231l,"大聰明",25));
        add(new User(31123l,"大智慧",27));
        add(new User(38751l,"張三",30));
    }
};
List<String> names = users.stream().map(User::getName).collect(Collectors.toList());
System.out.println(names.toString());
//distinct()過濾重復的姓名 count()計算數量
long count = names.stream().distinct().count();
System.out.println("去重后的個數:"+count);
if (names.size() == count){
    System.out.println("沒有重復");
}else {
    System.out.println("重復了，老鐵");
}