Question

我一直在追捕他们,以便这样做。

我想要做的是:

我有:

qualifiedCompetitors = new HashMap<String, List<String>>();

分配到散列图的数值是:

String[] events = {"Helicopter", "Amphibious", 
                     "4x4", "Fire Engine"};

String[][] competitors = 
{
   {"Greg", "Will", "Fleur", "Bill", "Bella", "Sally", "Olive", "Sal", "Dora", "Chas"}, 
   {"Yuri", "Abe", "Tim", "Fleur", "Bonnie", "Vera", "Ed", "Dan", "Jill", "Rose", "Zoe"}, 
   {"Tim", "Fleur", "Will",  "Bill", "Bella", "Dan", "Jill", "Rose", "Greg",  "Abe", "Sally", "Yuri", "Olive", "Sal", "Jim"}, 
   {"Jill", "Rose", "Greg", "Abe", "Bonnie", "Vera", "Ed", "Zoe", "Ben", "Freda", "Chuck", "Fred"}
};

关键筛选是事件的名称,清单是竞争者名单。我想建立一个所有竞争者的新名单。

因此,我尝试了诸如:

ListOfAllqualifiedCompetitors = new ArrayList(qualifiedCompetitors.values());

接着通过散列图进行:

Collection allCompetitors = qualifiedCompetitors.values();

System.out.println("Values of Collection created from Hashtable are :");
//iterate through the collection
Iterator itr = allDrivers.iterator();
while(itr.hasNext())
  System.out.println(itr.next());

看到我拿到什么东西,什么是罚款,但我不能再增加一个新名单? 我知道我没有什么明显的建议?

Answer 1

我的think <>/em>请你再说的是,你有一张<编码>String/code>至List<String>,你想积累所有<代码>。清单和编号;String> 列入一个较长的清单。如果是:

final Map<String, List<String>> eventsToCompetitors;
final List<String> allCompetitors;

eventsToCompetitors = ...

allCompetitors = new ArrayList<String>();
for (Collection<String> competitors : eventsToCompetitors.values())
{
  allCompetitors.addAll(competitors);
}

Answer 2

List<String> allCompetitors = new ArrayList<String>();
for(Set<String> competitors : qualifiedCompetitors.values()) {
    allCompetitors.addAll(competitors);
}

你们指的是什么?

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