一、去除List中重复的String

public List<String> removeStringListDupli(List<String> stringList) {
   Set<String> set = new LinkedHashSet<>();
   set.addAll(stringList);
   stringList.clear();
   stringList.addAll(set);
   return stringList;
}

或使用Java8的写法:

List<String> unique = list.stream().distinct().collect(Collectors.toList());

二、List中对象去重

比如现在有一个 Person类:

public class Person {
   private Long id;
   private String name;
   public Person(Long id, String name) {
       this.id = id;
       this.name = name;
   }
   public Long getId() {
       return id;
   }
   public void setId(Long id) {
       this.id = id;
   }
   public String getName() {
       return name;
   }
   public void setName(String name) {
       this.name = name;
   }
   @Override
   public String toString() {
       return "Person{" +
               "id=" + id +
               ", name='" + name + '\'' +
               '}';
   }
}

重写Person对象的equals()方法和hashCode()方法:

@Override
   public boolean equals(Object o) {
       if (this == o) return true;
       if (o == null || getClass() != o.getClass()) return false;
       Person person = (Person) o;
       if (!id.equals(person.id)) return false;
       return name.equals(person.name);
   }
   @Override
   public int hashCode() {
       int result = id.hashCode();
       result = 31 * result + name.hashCode();
       return result;
   }

下面对象去重的代码:

Person p1 = new Person(1l, "jack");
       Person p2 = new Person(3l, "jack chou");
       Person p3 = new Person(2l, "tom");
       Person p4 = new Person(4l, "hanson");
       Person p5 = new Person(5l, "胶布虫");
       List<Person> persons = Arrays.asList(p1, p2, p3, p4, p5, p5, p1, p2, p2);
       List<Person> personList = new ArrayList<>();
       // 去重
       persons.stream().forEach(
               p -> {
                   if (!personList.contains(p)) {
                       personList.add(p);
                   }
               }
       );
       System.out.println(personList);

List 的contains()方法底层实现使用对象的equals方法去比较的,其实重写equals()就好,但重写了equals最好将hashCode也重写了。


可以参见:http://stackoverflow.com/questions/30745048/how-to-remove-duplicate-objects-from-java-arraylist

http://blog.csdn.net/growing_tree/article/details/46622579


三、根据对象的属性去重


下面要根据Person对象的id去重,那该怎么做呢?

写一个方法吧:

 public static List<Person> removeDupliById(List<Person> persons) {
       Set<Person> personSet = new TreeSet<>((o1, o2) -> o1.getId().compareTo(o2.getId()));
       personSet.addAll(persons);
       return new ArrayList<>(personSet);
   }


通过Comparator比较器,比较对象属性,相同就返回0,达到过滤的目的。


再来看比较炫酷的Java8写法:


import static java.util.Comparator.comparingLong;

import static java.util.stream.Collectors.collectingAndThen;

import static java.util.stream.Collectors.toCollection;


// 根据id去重


    List<Person> unique = persons.stream().collect(collectingAndThen(toCollection(() -> new TreeSet<>(comparingLong(Person::getId))), ArrayList::new));


这段炫酷的代码是google的,还不明白是怎么个原理,等我好好研究一下,再专门写篇文章好好阐述一下。



 public static <T> Predicate<T> distinctByKey(Function<? super T, Object> keyExtractor) {
       Map<Object, Boolean> map = new ConcurrentHashMap<>();
       return t -> map.putIfAbsent(keyExtractor.apply(t), Boolean.TRUE) == null;
   }
// remove duplicate
       persons.stream().filter(distinctByKey(p -> p.getId())).forEach(p -> System.out.println(p));


java8 确实简化了很多冗长的操作,精简了代码,小伙,研究java8去吧!


博主目前用的:

	/**
	 * 
	 * distinctByKey:自定义拉姆达表达式去重
	 * 
	 * @param keyExtractor
	 * @return Predicate<T>
	 */
	public static <T> Predicate<T> distinctByKey(Function<? super T, Object> keyExtractor) {
		Map<Object, Boolean> seen = new ConcurrentHashMap<>();
		return object -> seen.putIfAbsent(keyExtractor.apply(object), Boolean.TRUE) == null;
	}


List<DeviceFactorVO> collect = queryDeviceList.stream().filter(distinctByKey(DeviceFactorVO::getPolluteCode)).collect(Collectors.toList());


参考:

http://www.cnblogs.com/jizha/p/java_arraylist_duplicate.html


http://stackoverflow.com/questions/29670116/remove-duplicates-from-a-list-of-objects-based-on-property-in-java-8


版权声明:取长补短,学以致用……


原文链接:http://blog.csdn.net/jiaobuchong/article/details/54412094

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