Question

I have a List<A> that contains some objects A with the attribute id. What I need is the list to contain only one object for an id.

public List<A> findAllA() {
    List<A> aList; // fetches the elements from db

    HashMap<Long, A> aMap = new HashMap<>();
    for (A a: aList) {
        aMap.put(a.getId(), a);
    }

    return new ArrayList<>(aMap.values()); 
}

So I put my list elements in a HashMap, because a HashMap holds distinct keys. Then a new ArrayList<A> with the values of the map is returned. This does work fine, but is there a more efficient way to achieve the result? Or at least a shorter one?

Answer 1

如果您要查找较少的字符/代码行，可以使用Stream s：

return aList.stream().distinct().collect(Collectors.toList());

这假设您的A班级会覆盖equals，并且A的两个实例等于且当且仅当它们具有相同的ID时。

当然，如果是这种情况，您也可以使用HashSet来消除重复项：

return new ArrayList<A>(new HashSet<A>(alist));

如果您的A课程未覆盖equals（或equals上述行为不正常），请使用HashMap，这是一个有效的线性时间解。你无法在平均运行时间方面做得更好。

Answer 2

使用Java 8流：

List<A> aList; // fetches the elements from db

Collection<A> original= aList.stream()
   .<Map<Integer, A>> collect(HashMap::new,(m,e)->m.put(e.getId(), e), Map::putAll).values();

Answer 3

如果equals()/hashCode()方法依赖于id，您只需从Set创建List：

Set<A> set = new HashSet<>(aList);

Set根据equals()删除重复的元素等于如果Set类型不合适，您仍可将其转换为List：

List<A> newList = new ArrayList<A>(new HashSet<A>(alist));

Answer 4

按照你的逻辑，你也可以尝试这样做

public List<A> findAllA() {
        List<A> aList; // fetches the elements from db

        HashMap<Long, A> aMap = new HashMap<>();
        List<Long> processedList = new ArrayList<>();
        for (A a: aList) {
            if(!processedList.contains(a.getId())) {
                processedList.add(a.getId());
                aMap.put(a.getId(), a);
            }
        }

        return new ArrayList<>(aMap.values());
    }

如何过滤掉列表<a> so that the List</a> <a> only contains elements of distinct A.getId()?

4 个答案: