Best way to sort list with index array

Question

I would like to sort a list of strings with an array of integers as the new indexes:

Integer[] newIndexes = {0,2,1};
List<String> fruits = new ArrayList<>();
fruits.add("Apple");
fruits.add("Pear");
fruits.add("Banana");
fruits.sort....

What would be the best way to do this?

Just use a TreeMap<Integer, String>, put index and string and then get the values sorted by index. — Thomas
– Thomas, Commented Nov 14, 2014 at 12:34

Keppil · Accepted Answer · 2014-11-14 12:36:47Z

3

Using Java 8, you can do it using a stream of the indexes and the map() method:

List<String> sortedFruits = Stream.of(newIndexes).map(fruits::get).collect(Collectors.toList());

gives the list

[Apple, Banana, Pear]

answered Nov 14, 2014 at 12:36

Keppil

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1 Comment

Meneer Venus Over a year ago

sadly enough i have to use Java 7, still +1

Meneer Venus · Accepted Answer · 2014-11-14 12:43:14Z

0

You can use a SortedMap to map the integers to the string values. This way it will sort the string values according to the natural ordering of the keys:

SortedMap<Integer, String> map = new TreeMap<Integer, String>();
for (int i = 0; i < newIndexes.length; i++) {
    map.put(newIndexes[i], fruits.get(i));
}

Collection<String> sortedFruits = map.values();

If you intend to sort the strings alphabetically, then you can just use Collections#sort().

edited Nov 14, 2014 at 12:43

Meneer Venus

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answered Nov 14, 2014 at 12:37

M A

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