方法1
List<String> words = Arrays.asList("a", "b", "c", "d", "a", "d");List<String> results = HashMultiset.create(words).entrySet().stream().filter(w -> w.getCount() > 1).map(Multiset.Entry::getElement).collect(Collectors.toList());System.out.println(results);
方法2
可以修改返回值 返回去重的集合
public static <E> List<E> getListDuplicateElements(List<E> list){List<E> words = list;Set<E> repeated = new HashSet<>();List<E> results = new ArrayList<>();for (E word : words) {if (!repeated.add(word)) {results.add(word);}}return results;}
方法3
导入guava依赖
<!-- /artifact/com.google.guava/guava --><dependency><groupId>com.google.guava</groupId><artifactId>guava</artifactId><version>31.0-jre</version></dependency>
List<String> words = Arrays.asList("a", "b", "c", "d", "a", "d");List<String> results = new ArrayList<>();for (Multiset.Entry<String> entry : HashMultiset.create(words).entrySet()) {if (entry.getCount() > 1) {results.add(entry.getElement());}}System.out.println(results);
Lambda 来使用guava
List<String> words = Arrays.asList("a", "b", "c", "d", "a", "d");List<String> results = HashMultiset.create(words).entrySet().stream().filter(w -> w.getCount() > 1).map(Multiset.Entry::getElement).collect(Collectors.toList());System.out.println(results);
方法4
public List getRepeatList(List list) {List list2 = new ArrayList();for (int i = 0; i < list.size(); i++) {for (int j = i + 1; j < list.size(); j++) {if (list.get(i) !="" && list.get(i).equals(list.get(j))) {list2.add(list.get(i));break;}}}return list2;}
/*** 获取list 集合重复元素** @param list* @param <E>* @return*/public static <E> List<E> getDuplicateElements(List<E> list) {return list.stream() .filter( i -> i!="") // list 对应的 Stream 并过滤"".collect(Collectors.toMap(e -> e, e -> 1, Integer::sum)) // 获得元素出现频率的 Map，键为元素，值为元素出现的次数.entrySet().stream() // 所有 entry 对应的 Stream.filter(e -> e.getValue() > 1) // 过滤出元素出现次数大于 1 (重复元素）的 entry.map(Map.Entry::getKey)// 获得 entry 的键（重复元素）对应的 Stream.collect(Collectors.toList()); // 转化为 List}
题外
//定义一个100元素的集合，包含A-ZList<String> list = new LinkedList<>();for (int i =0;i<100;i++){list.add(String.valueOf((char)('A'+Math.random()*('Z'-'A'+1))));}System.out.println(list);//统计集合重复元素出现次数，并且去重返回hashmapMap<String, Long> map = list.stream().collect(Collectors.groupingBy(Function.identity(),Collectors.counting()));System.out.println(map);//由于hashmap无序，所以在排序放入LinkedHashMap里(key升序)Map<String, Long> sortMap = new LinkedHashMap<>();map.entrySet().stream().sorted(paringByKey()).forEachOrdered(e -> sortMap.put(e.getKey(), e.getValue()));System.out.println(sortMap);//获取排序后map的key集合List<String> keys = new LinkedList<>();sortMap.entrySet().stream().forEachOrdered(e -> keys.add(e.getKey()));System.out.println(keys);//获取排序后map的value集合List<Long> values = new LinkedList<>();sortMap.entrySet().stream().forEachOrdered(e -> values.add(e.getValue()));System.out.println(values);
已逗号隔开返回字符串
String collect = name.stream().map(String::valueOf).collect(Collectors.joining(","));
过滤条件
List<Integer> collect = list.stream().filter(item -> item >= 1 && item <= 10).collect(Collectors.toList());
