I just finished the main part of the current data structures project, and am working on collecting the statistics. One requirement is that a count of all the references within the TreeMap be recorded.
This Map contains a 31,000+ nodes where a String is mapped to a TreeSet of indeterminate size. I need to traverse the map and keep a runnin开发者_JAVA百科g count of the number of items in the set.
Originally my idea was this:
Set<String> keySet= lyricWords.keySet();
Iterator<String> iter= keySet.iterator();
String current= iter.next();
while (iter.hasNext){
runCount+= lyricWords.get(current).size();
}
The runtime for this is far too long to be acceptable. Is there a more efficient way to do this on the final structure? I could keep a count as the map is built, but the professor wants the numbers to be based on the final structure itself.
I'm not sure. But, probably, you have infinitive loop. Try:
runCount+= iter.next().size();
for (Map.Entry<String, TreeSet> e: lyricWords.entrySet()) {
runCount+= e.getValue().size();
}
I dont see a problem with keeping a count as the map is built.
The count will be correct at the end, and you wont have to incur the cost of iterating through the entire thing again.
I think that the tree can and should keep track of its size
This isn't of much use to you since this is an assignment you're working on, but this is an example where a data structure specifically designed for mapping keys to multiple values shows how much better it is than a Map<T, Collection<V>>
.
Guava's Multimap
collection type keeps track of the total number of entries it contains, so if you were using a TreeMultimap<String, Foo>
rather than a TreeMap<String, TreeSet<Foo>>
you could just call multimap.size()
to get the number you're looking for.
By the way, the Multimap
implementations store a running total of the number of entries which is updated when entries are added to or removed from it. You might be able to do this by doing some fancy stuff with subclassing the TreeMap
and wrapping the TreeSet
s that are added to it, but it would be quite challenging to make it all work properly I think.
精彩评论