Sorting an array of String based on the strings sizes (length)

Question

How do I sort an an array of String based on the strings sizes i.e String#length?

possible duplicate of [Custom String Length Comparator: what's my mistake?](http://stackoverflow.com/questions/5925532/custom-string-length-comparator-whats-my-mistake) — Gábor Bakos, Jan 29 '15 at 12:23
Java 8: `Arrays.sort(strings, Comparator.comparing(String::length));`. — Marko Topolnik, Jan 29 '15 at 12:42
similar to http://stackoverflow.com/questions/8938235/java-sort-an-array — user7610, Mar 25 '17 at 12:15

Sufiyan Ghori · Answer 1 · 2015-02-08T14:32:05.043

3

In Java 8, this can be done in one line,

Arrays.sort(randomString, (s1,s2) -> Integer.compare(s1.length(), s2.length()));

If you want reverse order (higher-length to lower-length),

change it to,

Arrays.sort(randomString, (s1,s2) -> Integer.compare(s2.length(), s1.length()));

Another approach,

use Comparator.comparing(String::length),

Arrays.sort(yourArray, Comparator.comparing(String::length));

to reverse the order,

Arrays.sort(yourArray, Comparator.comparing(String::length).reversed());

edited Feb 08 '15 at 14:32

answered Jan 29 '15 at 13:04

Sufiyan Ghori

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1

Maybe a one-liner, but doesn't sort in-place and needlessly copies the data. – Marko Topolnik Jan 29 '15 at 13:06

Mnementh · Answer 2 · 2015-01-29T14:47:33.007

2

You can implement a Comparator that uses the length and use Arrays.sort with your Comparator. The Comparator could look like this:

class StringComparator implements Comparator<String>{
   public int compare(String o1, String o2){
      return Integer.compare(o1.length(), o2.length());
   }
}

Now you could sort with the following call:

Arrays.sort(strings, new StringComparator());

edited Jan 29 '15 at 14:47

answered Jan 29 '15 at 12:23

Mnementh

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Sorting an array of String based on the strings sizes (length)

2 Answers2