import java.util.AbstractList;
公共类ItemSet扩展了AbstractList {
private Item[] arr;
private ItemClass itemClass;
public ItemSet(Item item) {
arr = new Item[1];
arr[0] = item;
}
/*
* (non-Javadoc)
*
* @see java.util.AbstractList#add(java.lang.Object)
*/
@Override
public boolean add(Item e) {
boolean isNotAdded = true;
for (int i = 0; i < arr.length; i++) {
if (e.getRule().compareTo(arr[i].getRule())) {
if (e.getDot() == arr[i].getDot()) {
isNotAdded = false;
break;
}
}
}
if (isNotAdded) {
Item[] oldArr = arr;
arr = new Item[oldArr.length + 1];
System.arraycopy(oldArr, 0, arr, 0, oldArr.length);
arr[oldArr.length] = e;
}
return isNotAdded;
}
@Override
public Item get(int index) {
return arr[index];
}
@Override
public int size() {
return arr.length;
}
// SETTER
/**
* @param itemClass
* the itemClass to set
*/
public void setItemClass(ItemClass itemClass) {
this.itemClass = itemClass;
}
//
// GETTER
/**
* @return the itemClass
*/
public ItemClass getItemClass() {
return itemClass;
}
//
}
如何覆盖例如删除方法? 如何不使用自己的列表实现?
答案 0 :(得分:3)
如果您不想在界面中实现其他方法,只需定义它们以引发异常:
public bool remove(Object o)
{
throw new UnsupportedOperationException();
}
如果你不想这样做,你应该继承一个类,而不是实现一个接口。
答案 1 :(得分:-1)
如果您不想继承remove(),请不要扩展AbstractList
,而是将您想要的方法委派给ArrayList
。
喜欢这样(并且为了上帝的爱,不要评论你的二传手和吸气鬼)
public class ItemSet {
private final ArrayList<Item> items = new ArrayList<Item>();
private ItemClass itemClass;
public ItemSet(Item item) {
items.add(item);
}
public boolean add(Item e) {
boolean isNotAdded = true;
for (Item item : items)
if (e.getRule().compareTo(item.getRule()))
if (e.getDot() == item.getDot()) {
isNotAdded = false;
break;
}
// why are you adding here?
if (isNotAdded)
items.add(e);
return isNotAdded;
}
public Item get(int index) {
return items.get(index);
}
public int size() {
return items.size();
}
public void setItemClass(ItemClass itemClass) {
this.itemClass = itemClass;
}
public ItemClass getItemClass() {
return itemClass;
}
}
}