【发布时间】:2016-12-23 11:22:20
【问题描述】:
我必须使用 ArrayList 创建
HashMap<Integer,List<StoreItem>>。我想要 HashSet 基于商店编号,就像我们在 Ios 中所做的那样,例如myArrayList.distinctUnionOfObjects.storeNumber
public static List<StoreItem> getStoreList(){
List<StoreItem> items = new ArrayList<>();
items.add(new StoreItem(503,"Brufeen",10));
items.add(new StoreItem(503,"hydriline",8));
items.add(new StoreItem(503,"Brufeen",10));
items.add(new StoreItem(503,"capsule",2));
items.add(new StoreItem(503,"Disprin",6));
items.add(new StoreItem(504,"Pixlar",9));
items.add(new StoreItem(504,"Luprin",17));
items.add(new StoreItem(504,"BOOM BOOM",14));
items.add(new StoreItem(504,"Glucophage",22));
items.add(new StoreItem(504,"Panadol",16));
items.add(new StoreItem(549,"Pixlar",9));
items.add(new StoreItem(549,"Luprin",17));
items.add(new StoreItem(549,"BOOM BOOM",14));
items.add(new StoreItem(549,"Glucophage",22));
items.add(new StoreItem(549,"Panadol",16));
return items;
}
public static HashMap<Integer,List<StoreItem>> getStoreHashMap(){
List<StoreItem> oldStores = StoreFactory.getStoreList();
Set<StoreItem> uniqueSet = new HashSet<StoreItem>(oldStores);
HashMap<Integer,List<StoreItem>> dictionaryOfStore = new HashMap<>();
for(StoreItem keyItem : uniqueSet )
{
List<StoreItem> storeInSection = new ArrayList<>();
for(StoreItem oldItem : oldStores)
{
if(keyItem.getStoreNumber() == oldItem.getStoreNumber()){
storeInSection.add(oldItem);
}
}
dictionaryOfStore.put(keyItem.getStoreNumber(),storeInSection);
}
return dictionaryOfStore;
}
uniqueSet 返回 15,因为它比较了整个 StoreItem 对象。我需要一种基于 StoreItem.getStoreNumber()
创建 HashSet 的方法StoreItem 在哪里
public class StoreItem {
private int storeNumber;
private String capsuleNames;
private int quantity;
public StoreItem(int storeNumber, String capsuleNames, int quantity) {
this.storeNumber = storeNumber;
this.capsuleNames = capsuleNames;
this.quantity = quantity;
}
}
或者有没有办法覆盖哈希集比较。
而且我知道我可以通过 For 循环中的检查来做到这一点。
【问题讨论】:
-
我在考虑两个选项:(1) 使用
TreeSet和自定义比较器而不是HashSet(2) 如果您坚持使用HashSet,请将其设为@ 987654328@ 其中MyList扩展ArrayList与适当的equals()和hashCode()。