fastest way to sort a key/value list in java

Question

I got the following situation:

I have a list of unsorted keys and a different list of String terms. Both are connected in a way, meaning it looks kinda like this:

1 contents
5 term
2 queue
etc.

This list now contains 1000+ entries and I want to find a fast way to sort these two lists in descending order but of course I need the persistent connection to the strings.

5 term
2 queue
1 contents

I thought about putting the keys and values in a TreeMap, but the thing is, there can be duplicate keys, and I want to preserve them. Any ideas on this?

SJuan76 · Accepted Answer · 2013-01-28 16:37:04Z

2

Create a class that holds the pair of values.

Make it extend Comparable.

Implement compareTo() so it orders by the number.

Use any of the sorting methods already in the API (Arrays.sort, or a SortedSet).

Have fun.

answered Jan 28, 2013 at 16:37

SJuan76

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4 Comments

mikey Over a year ago

Using this approach PogoMips can simply use a List, correct, no need for the TreeMap?

SJuan76 Over a year ago

@mikey Yes, no need of a Map. The pairs are kept together because you only deal with the object that encapsulates them.

PogoMips Over a year ago

hmm.. so it works but I'm not really convinced. It takes about 7 seconds to create 1,000 objects and put them in a list. And I'm going to have lists with 10,000+ entries.. are there any other options?

SJuan76 Over a year ago

If you know the dimensions a priori, an array should be faster than the usual list. ArrayList has a constructor that accepts an initial capacity too. Other than that, improvements start being more complicated and suited for particular scenarios.

AlexR · Accepted Answer · 2013-01-28 16:39:55Z

0

Use TreeMap with IntegerComparator implemented like the following:

class IntegerComparator implements Comparator<Integer> {
   public int compare(Integer one, Integer two) {
        return two - one;
   }
}

answered Jan 28, 2013 at 16:39

AlexR

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2 Comments

jarnbjo Over a year ago

Not if there are duplicate "keys" (or sorting indexes).

SJuan76 Over a year ago

@jarnbjo, duplicate keys only requires that in the map, instead of storing <Integer, String> you store <Integer, List<String>>

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