如何优雅的找出ArrayList中的重复元素？

Question

故意强调了优雅，所以双循环什么的就别来了..先谢为敬

Answer 1

可以参考：

    public static void main(String[] args) {
        List<String> list = new ArrayList<String>();
        list.add("string1");
        list.add("string1");
        list.add("string1");
        list.add("string1");
        list.add("string3");
        list.add("string2");
        list.add("string2");
        HashMap<String, Integer> hashMap = new HashMap<String, Integer>();
        for (String string : list) {
            if (hashMap.get(string) != null) {
                Integer value = hashMap.get(string);
                hashMap.put(string, value+1);
                System.out.println("the element:"+string+" is repeat");
            } else {
                hashMap.put(string, 1);
            }
        }
    }

Answer 2

如果你用上了 Java8，使用 Stream 来实现这个功能（
getDuplicateElements 方法）会很优雅：

public class DemoCode {

    public static void main(String[] args) {
        List<String> list = Arrays.asList("a", "b", "c", "d", "a", "a", "d", "d");
        List<String> duplicate = getDuplicateElements(list);

        System.out.println("list 中重复的元素：" + duplicate);
    }

    public static <E> List<E> getDuplicateElements(List<E> list) {
        return list.stream()                              // list 对应的 Stream
                   .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
    }
}

运行结果：

Answer 3

List<String> list = new ArrayList<String>();
list.add("aaa");
list.add("bbb");
list.add("ccc");
list.add("ddd");
list.add("aaa");
list.add("aaaa");
list.add("eee");
list.add("bbb");
list.add("ccc");
StringBuilder builder = new StringBuilder();
for(String str : list) {
    // 如果不存在返回 -1。
    if(builder.indexOf(","+str+",") > -1) {
        System.out.println("重复的有："+str);
    } else {
        builder.append(",").append(str).append(",");
    }
}

Answer 4

用 Guava 可以这样写

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);

With Lambda

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);

不使用 Guava

List<String> words = Arrays.asList("a", "b", "c", "d", "a", "d");
Set<String> repeated = new HashSet<>();
List<String> results = new ArrayList<>();
for (String word : words) {
  if (!repeated.add(word)) {
    results.add(word);
  }
}
System.out.println(results);

Answer 5

优雅的话当然是Java8的新特性了

Answer 6

试试小弟这个吧，简洁好用
不需要转map判断 ：)
list.stream().filter(i -> list.stream().filter(i::equals).count()>1).collect(Collectors.toSet());

Answer 7

List<String> originList; //初始化和判空不写了
List<String> distinctList = originList.stream().distinct().collect(Collectors.toList()).size()
if(originList.size() != distinctList.size()){
    //存在重复元素
}

Answer 8

//multiSet允许重复，并且可以求某个元素重复个数
        ArrayList arrayList = Lists.newArrayList("1","2","2","3","4","1");
        HashMultiset set = HashMultiset.create(arrayList);
        //求出哪个重复元素
        Set obj = (Set)set.stream().filter(d->set.count(d)>1).collect(Collectors.toSet());
        System.out.println(obj);

Answer 9

优不优雅，只是别人封装罢了。