Sort ArrayList of strings by length difference

Question

I want to order an ArrayList of strings by length, but not just in numeric order.

Say for example, the list contains these words:

cucumber
aeronomical
bacon
tea
telescopic
fantasmagorical

They need to be ordered by their difference in length to a special string, for example:

intelligent

So the final list would look like this (difference in brackets):

aeronomical     (0)
telescopic      (1)
fantasmagorical (3) - give priority to positive differences? doesn't really matter
cucumber        (3)
bacon           (6)
tea             (8)

As noted in a link-only answer, this can be accomplished with Collections.sort(), which takes an explicit Comparator — Trenton McKinney
– Trenton McKinney, Commented Oct 30, 2020 at 2:57

Barend · Accepted Answer · 2011-09-27 21:18:18Z

30

Use a custom comparator:

public class MyComparator implements java.util.Comparator<String> {

    private int referenceLength;

    public MyComparator(String reference) {
        super();
        this.referenceLength = reference.length();
    }

    public int compare(String s1, String s2) {
        int dist1 = Math.abs(s1.length() - referenceLength);
        int dist2 = Math.abs(s2.length() - referenceLength);

        return dist1 - dist2;
    }
}

Then sort the list using java.util.Collections.sort(List, Comparator).

answered Sep 27, 2011 at 21:18

Barend

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2 Comments

Stephen Denne Over a year ago

Priority to positive differences can be given by multiplying positive differences by two, and negative differences by negative two, then adding one.

Steve Kuo Over a year ago

You should make referenceLength final.

SamHoque · Accepted Answer · 2018-09-21 06:51:09Z

17

If you are using java 8 you can also try using this lambda

packages.sort(Comparator.comparingInt(String::length));

answered Sep 21, 2018 at 6:51

SamHoque

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Comments

Romano Zumbé · Accepted Answer · 2017-07-03 13:05:55Z

13

If you're using Java 8+ you can use a lambda expression to implement (@Barend's answer as) the comparator

List<String> strings = Arrays.asList(new String[] {"cucumber","aeronomical","bacon","tea","telescopic","fantasmagorical"});
strings.sort((s1, s2) -> Math.abs(s1.length() - "intelligent".length()) - Math.abs(s2.length() - "intelligent".length()));

edited Jul 3, 2017 at 13:05

Romano Zumbé

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answered Aug 1, 2014 at 1:41

Josh Johnson

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2 Comments

Mugoma J. Okomba Over a year ago

Best solution for Java 8+. Short and precise.

PlsWork Over a year ago

In case you just want to sort by length: strings.sort((s1, s2) -> s1.length() - s2.length());

Trenton McKinney · Accepted Answer · 2020-10-30 02:59:45Z

5

String in Ascending order

class StringLengthListSort implements Comparator<String>{

    @Override
    public int compare(String s1, String s2) {
    return s1.length() - s2.length();
    }

    /**
     * @param args
     */
    public static void main(String[] args) {
    List<String> list = new ArrayList<String>();
    StringLengthListSort ss = new StringLengthListSort();
    list.add("ram");
    list.add("rahim");
    list.add("ramshyam");
    Collections.sort(list, ss);
    System.out.println(list);
    }

}

edited Oct 30, 2020 at 2:59

Trenton McKinney

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answered Aug 6, 2013 at 11:14

Krishna Kumar Chourasiya

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Comments

parsifal · Accepted Answer · 2011-09-27 21:19:22Z

3

You'd do this with the version of Collections.sort() that takes an explicit Comparator.

answered Sep 27, 2011 at 21:19

parsifal

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Comments

xyzzxy123 · Accepted Answer · 2020-08-08 07:00:53Z

3

I have a similar problem solved by lambda expression:

listBeforeSorting.sort((s1, s2) -> s1.length() - s2.length());

This way, we will get sorted-by-length(ascending order) list.

answered Aug 8, 2020 at 7:00

xyzzxy123

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Comments

Kaushal Dokania · Accepted Answer · 2021-05-24 16:55:03Z

3

Collections.sort(list, (a, b)->Integer.compare(a.length(), b.length()));

answered May 24, 2021 at 16:55

Kaushal Dokania

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Comments

Chris Mart · Accepted Answer · 2018-08-28 17:49:00Z

2

The use of a custom comparator is correct. This is one way to implement it:

    Comparator c = new Comparator<String>()
    {
        public int compare(String s1, String s2) {
            return Integer.compare(s1.length(), s2.length());
        }
    };
    Collections.sort(results, c);
    return results;

edited Aug 28, 2018 at 17:49

answered Aug 28, 2018 at 17:43

Chris Mart

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Comments

Ross Bu · Accepted Answer · 2019-10-10 19:42:43Z

0

simple with Java8 solution with Comparator and Method reference only: Stream.of(list).flatMap(Collection::stream).sorted( Comparator.comparing( String::length)).collect(toList());

answered Oct 10, 2019 at 19:42

Ross Bu

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Comments

Ashish Kumar · Accepted Answer · 2020-10-29 15:45:10Z

0

The shortest code for this-

public static void main(String... str) {
        List.of("am", "I", "Best", "the").stream().sorted((a, b) -> a.length() - b.length())
                .forEach(System.out::println);
    }

answered Oct 29, 2020 at 15:45

Ashish Kumar

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Comments

Taz · Accepted Answer · 2021-02-13 02:23:39Z

-1

List<String> list = Arrays.asList("geeksforgeeks", "geeksfor", "geeks");
Collections.sort(list, new Comparator<String>(){
       public int compare(String s1, String s2){
                return s1.length() - s2.length();
       }
});

answered Feb 13, 2021 at 2:23

Taz

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Comments

Procrastinator · Accepted Answer · 2023-04-17 09:08:27Z

-1

Sort the list of string by length using java8

String[] strArr = { "Apples","CSS", "HTML", "Oracle", "Dart"};
    
List<String> result = Arrays.asList(strArr).stream().sorted( (str1,str2) ->  str1.length() - str2.length()).collect(Collectors.toList());
    
System.out.println(result); // output in Ascending Order : [CSS, HTML, Dart, Apples, Oracle]

edited Apr 17, 2023 at 9:08

Procrastinator

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answered Apr 10, 2023 at 14:17

Amit Thakur

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