是否存在集合中的任何一个成员都可以解决的集合？

Question

我试图在 Java（或 Groovy）中找到一个数据结构，它可以像这样工作：

MemberAdressableSetsSet mass = new MemberAdressableSetsSet();

mass.addSet(["a","b"]);

mass.addSet(["c","d","e"]);

mass.get("d").add("f");

String output = Arrays.toString(mass.get("e").toArray());

System.out.println(output); // [ "c", "d", "e", "f" ] (ordering irrelevant)

有没有这样的东西？如果没有，有没有办法用普通的 Java 代码来实现这样的东西，而不会给 CPU 或数周的内存带来噩梦？

编辑：更严格

MemberAdressableSetsSet mass = new MemberAdressableSetsSet();

Set<String> s1 = new HashSet<String>();

s1.add("a");

Set<String> s2 = new HashSet<String>();

s2.add("c");s2.add("d");s2.add("e");

mass.addSet(s1);

mass.addSet(s2);

Set<String> s3 = new HashSet<String>();

s3.add("a");s3.add("z");

mass.addSet(s3);

/* s3 contains "a", which is already in a subset of mass, so:

 * Either

 *   - does nothing and returns false or throws Exception

 *   - deletes "a" from its previous subset before adding s3

 *      => possibly returns the old subset

 *      => deletes the old subset if that leaves it empty

 *      => maybe requires an optional parameter to be set

 *   - removes "a" from the new subset before adding it

 *      => possibly returns the new subset that was actually added

 *      => does not add the new subset if purging it of overlap leaves it empty

 *      => maybe requires an optional parameter to be set

 *   - merges all sets that would end up overlapping

 *   - adds it with no overlap checks, but get("a") returns an array of all sets containing it

 */

mass.get("d").add("f");

String output = Arrays.toString(mass.get("e").toArray());

System.out.println(output); // [ "c", "d", "e", "f" ] (ordering irrelevant)

mass.get("d")将返回Set<T>包含mass. "d"类似于 get() 的工作方式，例如HashMap：

HashMap<String,LinkedList<Integer>> map = new HashMap<>();

LinkedList<Integer> list = new LinkedList<>();

list.add(9);

map.put("d",list);

map.get("d").add(4);

map.get("d"); // returns a LinkedList with contents [9,4]

Answer 1

到目前为止我能想到的最好的看起来像这样：import java.util.HashMap;import java.util.Set;public class MemberAdressableSetsSet {&nbsp; &nbsp; private int next_id = 1;&nbsp; &nbsp; private HashMap<Object,Integer> members = new HashMap();&nbsp; &nbsp; private HashMap<Integer,Set> sets = new HashMap();&nbsp; &nbsp; public boolean addSet(Set s) {&nbsp; &nbsp; &nbsp; &nbsp; if (s.size()==0) return false;&nbsp; &nbsp; &nbsp; &nbsp; for (Object member : s) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if (members.get(member)!=null) return false;&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; sets.put(next_id,s);&nbsp; &nbsp; &nbsp; &nbsp; for (Object member : s) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; members.put(member,next_id);&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; next_id++;&nbsp; &nbsp; &nbsp; &nbsp; return true;&nbsp; &nbsp; }&nbsp; &nbsp; public boolean deleteSet(Object member) {&nbsp; &nbsp; &nbsp; &nbsp; Integer id = members.get(member);&nbsp; &nbsp; &nbsp; &nbsp; if (id==null) return false;&nbsp; &nbsp; &nbsp; &nbsp; Set set = sets.get(id);&nbsp; &nbsp; &nbsp; &nbsp; for (Object m : set) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; members.remove(m);&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; sets.remove(id);&nbsp; &nbsp; &nbsp; &nbsp; return true;&nbsp; &nbsp; }&nbsp; &nbsp; public boolean addToSet(Object member, Object addition) {&nbsp; &nbsp; &nbsp; &nbsp; Integer id = members.get(member);&nbsp; &nbsp; &nbsp; &nbsp; if (id==null) throw new IndexOutOfBoundsException();&nbsp; &nbsp; &nbsp; &nbsp; if (members.get(addition)!=null) return false;&nbsp; &nbsp; &nbsp; &nbsp; sets.get(id).add(addition);&nbsp; &nbsp; &nbsp; &nbsp; members.put(addition,id);&nbsp; &nbsp; &nbsp; &nbsp; return true;&nbsp; &nbsp; }&nbsp; &nbsp; public boolean removeFromSet(Object member) {&nbsp; &nbsp; &nbsp; &nbsp; Integer id = members.get(member);&nbsp; &nbsp; &nbsp; &nbsp; if (id==null) return false;&nbsp; &nbsp; &nbsp; &nbsp; Set s = sets.get(id);&nbsp; &nbsp; &nbsp; &nbsp; if (s.size()==1) sets.remove(id);&nbsp; &nbsp; &nbsp; &nbsp; else s.remove(member);&nbsp; &nbsp; &nbsp; &nbsp; members.remove(member);&nbsp; &nbsp; &nbsp; &nbsp; return true;&nbsp; &nbsp; }&nbsp; &nbsp; public Set getSetClone(Object member) {&nbsp; &nbsp; &nbsp; &nbsp; Integer id = members.get(member);&nbsp; &nbsp; &nbsp; &nbsp; if (id==null) throw new IndexOutOfBoundsException();&nbsp; &nbsp; &nbsp; &nbsp; Set copy = new java.util.HashSet(sets.get(id));&nbsp; &nbsp; &nbsp; &nbsp; return copy;&nbsp; &nbsp; }}这有一些缺点：集合不可直接访问，这使得所有Set未通过显式定义的转换方法公开的方法和属性都不可访问，除非克隆是可接受的选项类型信息丢失。说Set<Date>加了一个。它不会抱怨试图添加，例如，一个File对象到那个集合。至少丢失的 Sets 类型信息没有扩展到它们的成员：它们Set.contains()仍然按预期工作，尽管双方Object在被比较之前都被类型转换为contains()。因此，当被问及是否包含时，包含的集合(Object)3不会返回 true&nbsp;(Object)3L，反之亦然。

Answer 2

您需要多久访问一个元素？可能值得使用地图并将相同的Set引用存储在多个键下。我会防止地图和子集的外部突变，并提供辅助方法来完成所有更新：public class MemberAdressableSets<T> {&nbsp; &nbsp; Map<T, Set<T>> data = new HashMap<>();&nbsp; &nbsp; public void addSet(Set<T> dataSet) {&nbsp; &nbsp; &nbsp; &nbsp; if (dataSet.stream().anyMatch(data::containsKey)) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; throw Exception("Key already in member addressable data");&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; Set<T> protectedSet = new HashSet<>(dataSet);&nbsp; &nbsp; &nbsp; &nbsp; dataSet.forEach(d -> data.put(d, protectedSet));&nbsp; &nbsp; }&nbsp; &nbsp; public void updateSet(T key, T... newData) {&nbsp; &nbsp; &nbsp; &nbsp; Set<T> dataSet = data.get(key);&nbsp; &nbsp; &nbsp; &nbsp; Arrays.stream(newData).forEach(dataSet::add);&nbsp; &nbsp; &nbsp; &nbsp; Arrays.stream(newData).forEach(d -> data.put(d, dataSet));&nbsp; &nbsp; }&nbsp; &nbsp; public Set<T> get(T key) {&nbsp; &nbsp; &nbsp; &nbsp; return Collections.unmodifiableSet(data.get(key));&nbsp; &nbsp; }}或者，如果密钥不存在，您可以更新addSet和updateSet创建新实例并创建never 。您还需要扩展此类以处理合并集的情况。即处理用例：SetupdateSetthrowmass.addSet(["a","b"]);mass.addSet(["a","c"]);

Answer 3

该解决方案允许诸如mass.get("d").add("f");影响存储在 中的子集之类的事情mass，但有很大的缺点。import java.util.Iterator;import java.util.LinkedHashSet;import java.util.Set;public class MemberAdressableSetsSetDirect {&nbsp; &nbsp; private LinkedHashSet<Set> sets;&nbsp; &nbsp; public void addSet(Set newSet) {&nbsp; &nbsp; &nbsp; &nbsp; sets.add(newSet);&nbsp; &nbsp; }&nbsp; &nbsp; public Set removeSet(Object member) {&nbsp; &nbsp; &nbsp; &nbsp; Iterator<Set> it = sets.iterator();&nbsp; &nbsp; &nbsp; &nbsp; while (it.hasNext()) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; Set s = it.next();&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if (s.contains(member)) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; it.remove();&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; return s;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; return null;&nbsp; &nbsp; }&nbsp; &nbsp; public int removeSets(Object member) {&nbsp; &nbsp; &nbsp; &nbsp; int removed = 0;&nbsp; &nbsp; &nbsp; &nbsp; Iterator<Set> it = sets.iterator();&nbsp; &nbsp; &nbsp; &nbsp; while (it.hasNext()) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; Set s = it.next();&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if (s.contains(member)) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; it.remove();&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; removed++;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; return removed;&nbsp; &nbsp; }&nbsp; &nbsp; public void deleteEmptySets() {&nbsp; &nbsp; &nbsp; &nbsp; sets.removeIf(Set::isEmpty);&nbsp; &nbsp; }&nbsp; &nbsp; public Set get(Object member) {&nbsp; &nbsp; &nbsp; &nbsp; for (Set s : sets) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if (s.contains(member)) return s;&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; return null;&nbsp; &nbsp; }&nbsp; &nbsp; public Set[] getAll(Object member) {&nbsp; &nbsp; &nbsp; &nbsp; LinkedHashSet<Set> results = new LinkedHashSet<>();&nbsp; &nbsp; &nbsp; &nbsp; for (Set s : sets) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if (s.contains(member)) results.add(s);&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; return (Set[]) results.toArray();&nbsp; &nbsp; }}没有针对重叠的内置保护，因此我们的访问不可靠，并且引入了无数空集的可能性，这些空集需要通过手动调用定期清除，因为此解决方案无法检测子集是否被deleteEmptySets()修改直接访问。MemberAdressableSetsSetDirect massd = new MemberAdressableSetsSetDirect();Set s1 = new HashSet();Set s2 = new HashSet();Set s3 = new HashSet();s1.add("a");s1.add("b");s2.add("c");s2.add("d");s3.add("e");massd.addSet(s1);massd.addSet(s2);massd.get("c").add("a");// massd.get("a") will now either return the Set ["a","b"] or the Set ["a","c","d"]// (could be that my usage of a LinkedHashSet as the basis of massd//&nbsp; at least makes it consistently return the set added first)massd.get("e").remove("e");// the third set is now empty, can't be accessed anymore,// and massd has no clue about that until it's told to look for empty setsmassd.get("c").remove("d");massd.get("c").remove("c");// if LinkedHashSet makes this solution act as I suspected above,// this makes the third subset inaccessible except via massd.getAll("a")[1]此外，此解决方案也无法保留类型信息。这甚至不会发出警告：MemberAdressableSetsSetDirect massd = new MemberAdressableSetsSetDirect();Set<Long> s = new HashSet<Long>();s.add(3L);massd.addSet(s);massd.get(3L).add("someString");// massd.get(3L) will now return a Set with contents [3L, "someString"]