java实现了一个包含不同值的TransformList？
我试图创建TransformList的一个实现，它维护一个源列表之外的不同值列表。然而，对于实现应该如何将不同的值添加到我的hashmap和内部包含的不同列表中，我有点困惑。不过我觉得我的ListChangeListener.change应该行得通。但是我如何截取任何新的或删除的不同值，并将它们添加/删除到不同的映射和列表中呢
public class DistinctList<E> extends TransformationList<E,E> {
private final ObservableList<E> distinctList = FXCollections.observableArrayList();
private final ConcurrentHashMap<E,E> distinctValues = new ConcurrentHashMap<>();
private final ObservableList<E> source;
public DistinctList(ObservableList<E> source) {
super(source);
this.source = source;
source.stream().filter(s -> attemptAdd(s)).forEach(s -> distinctList.add(s));
}
private boolean attemptAdd(E e) {
final boolean result = distinctValues.putIfAbsent(e,e) == null;
if (result) {
distinctList.add(e);
}
return result;
}
private boolean attemptRemove(E e) {
final boolean result = distinctValues.remove(e, e);
if (result) {
distinctList.remove(e);
}
return result;
}
@Override
protected void sourceChanged(ListChangeListener.Change<? extends E> c) {
fireChange(new ListChangeListener.Change<E>(this) {
@Override
public boolean wasAdded() {
if (c.getAddedSubList().stream().filter(v -> distinctValues.contains(v) == false).findAny().isPresent()) {
return true;
}
else {
return false;
}
}
@Override
public boolean wasRemoved() {
if (c.getRemoved().stream().filter(v -> !source.contains(v)).findAny().isPresent()) {
return true;
}
else {
return false;
}
}
@Override
public boolean wasPermutated() {
return false;
}
@Override
protected int[] getPermutation() {
throw new AssertionError("getPermutation() not implemented");
}
@Override
public List<E> getRemoved() {
return c.getRemoved().stream().filter(v -> !source.contains(v)).collect(Collectors.toList());
}
@Override
public int getFrom() {
return 0;
}
@Override
public int getTo() {
return 0;
}
@Override
public boolean next() {
return c.next();
}
@Override
public void reset() {
c.reset();
}
});
}
@Override
public int getSourceIndex(int index) {
return IntStream.range(0,source.size()).filter(i -> source.get(i).equals(this.get(i))).findAny().orElse(-1);
}
@Override
public E get(int index) {
return distinctList.get(index);
}
@Override
public int size() {
return distinctList.size();
}
}
更新
我一直在研究这个问题，我想我找到了在哪里使用不同的值映射和列表来交互源代码更改。但是，当我的源列表删除一个值（以及其他具有相同hashcode/equals的值仍然存在）时，它会错误地从不同的值中删除该值。我做错了什么
public class DistinctList<E> extends TransformationList<E,E> {
private final ObservableList<E> distinctList = FXCollections.observableArrayList();
private final ConcurrentHashMap<E,E> distinctValues = new ConcurrentHashMap<>();
private final ObservableList<E> source;
public DistinctList(ObservableList<E> source) {
super(source);
this.source = source;
source.stream().forEach(s -> attemptAdd(s));
}
private boolean attemptAdd(E e) {
final boolean result = distinctValues.putIfAbsent(e,e) == null;
if (result) {
distinctList.add(e);
}
return result;
}
private boolean attemptRemove(E e) {
final boolean result = distinctValues.remove(e, e);
if (result) {
distinctList.remove(e);
}
return result;
}
@Override
protected void sourceChanged(ListChangeListener.Change<? extends E> c) {
ListChangeListener.Change<E> change = new ListChangeListener.Change<E>(this) {
@Override
public boolean wasAdded() {
if (c.getAddedSubList().stream().filter(v -> source.contains(v)).findAny().isPresent()) {
return true;
}
else {
return false;
}
}
@Override
public boolean wasRemoved() {
if (c.getRemoved().stream().filter(v -> source.contains(v) == false).findAny().isPresent()) {
return true;
}
else {
return false;
}
}
@Override
public boolean wasPermutated() {
return false;
}
@Override
protected int[] getPermutation() {
throw new AssertionError("getPermutation() not implemented");
}
@Override
public List<E> getRemoved() {
return c.getRemoved().stream().filter(v -> source.contains(v) == false)
.collect(Collectors.toList());
}
@Override
public int getFrom() {
return 0;
}
@Override
public int getTo() {
return 0;
}
@Override
public boolean next() {
return c.next();
}
@Override
public void reset() {
c.reset();
}
};
while (c.next()) {
if (c.wasAdded()) {
c.getAddedSubList().stream().filter(v -> !distinctValues.containsKey(v)).peek(a -> System.out.println("ADDING FROM MAP " + a)).forEach(a -> attemptAdd(a));
}
if (c.wasRemoved()) {
c.getRemoved().stream().filter(v -> distinctValues.containsKey(v)).peek(a -> System.out.println("REMOVING FROM MAP " + a)).forEach(a -> attemptRemove(a));
}
}
fireChange(change);
}
@Override
public int getSourceIndex(int index) {
return IntStream.range(0,source.size()).filter(i -> source.get(i).equals(this.get(i))).findAny().orElse(-1);
}
@Override
public E get(int index) {
return distinctList.get(index);
}
@Override
public int size() {
return distinctList.size();
}
}
# 1 楼答案
我想我明白了。如果我遗漏了什么，请告诉我