How to sort String array by length using Arrays.sort()

Question

I am trying to sort an array of strings according to their length using Arrays.sort(), but this sorts the strings lexicographically rather than by length. Here is my code:

S = "No one could disentangle correctly"
String W[] = S.split(" ");
Arrays.sort(W);

After sorting :

correctly
could
disentangle
no
one

but what I want is

no  //length = 2
one //length = 3
could //length = 4 and likewise
correctly
disentangle

How can I get the above output? Please give answer for JDK 1.7 & JDK1.8.

@matt So u mean I can not achieve this through Arrays.sort() — Zohra Khan
– Zohra Khan, Commented Mar 8, 2016 at 11:34
No I mean to use the 2 argument version of Arrays.sort(Object[], Comparator). — matt
– matt, Commented Mar 8, 2016 at 11:36
Arrays.sort() sorted the strings alphabetically. What is the sort criteria you wanted it to use instead? That's the part you will need to write your own comparator for. — Jane Nicholson
– Jane Nicholson, Commented Mar 8, 2016 at 11:36
Are you sorting by length? This answer may help: stackoverflow.com/questions/8632857/… — Jane Nicholson
– Jane Nicholson, Commented Mar 8, 2016 at 11:38

matt · Accepted Answer · 2020-01-21 07:23:04Z

49

For java 8 and above

 Arrays.sort(W, (a, b)->Integer.compare(a.length(), b.length()));

A more concise way is to use Comparator.comparingInt from Mano's answer here.

edited Jan 21, 2020 at 7:23

answered Mar 8, 2016 at 11:36

matt

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Comments

Manos Nikolaidis · Accepted Answer · 2016-03-08 11:40:41Z

37

Alternative to and slightly simpler than matt's version

Arrays.sort(W, Comparator.comparingInt(String::length));

answered Mar 8, 2016 at 11:40

Manos Nikolaidis

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3 Comments

matt Over a year ago

This could also open up using more complete comparators. Comparator.comparingInt(String::length).thenComparing(Comparator.naturalOrder());

Anurag Awasthi Over a year ago

How to get it in descending order?

michaeak Over a year ago

Comparator.comparingInt(String::length).reversed()

mmuzahid · Accepted Answer · 2016-03-08 11:49:18Z

19

If you are using JDK 1.8 or above then you could use lambda expression like matt answer. But if you are using JDK 1.7 or earlier version try to write a custom Comparator like this:

String S = "No one could disentangle correctly";
String W[] = S.split(" ");
Arrays.sort(W, new java.util.Comparator<String>() {
    @Override
    public int compare(String s1, String s2) {
        // TODO: Argument validation (nullity, length)
        return s1.length() - s2.length();// comparision
    }
});

answered Mar 8, 2016 at 11:49

mmuzahid

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Comments

Suraj Rao · Accepted Answer · 2019-01-31 09:49:19Z

5

Simply changing the parameter position we can sort in descending order.

Arrays.sort(W, (a,b)->b.length() - a.length());

edited Jan 31, 2019 at 9:49

Suraj Rao

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answered Jan 31, 2019 at 9:45

Abhijeet Gupta

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Comments

Zabuzard · Accepted Answer · 2020-01-21 07:17:12Z

Java 8 is pretty easy. as mentioned in the above answer. I was solving some problems on hackerrank and there they are using java 7. so I had to write the java7. That is selection sort. though time complexity is not great O(n^2) however it serves the purpose.

public static void main(String[] args) {
    // I am taking list and again converting to array. please ignore. you can directly take array of          string and apply the logic.
        List<String> listOfString = new ArrayList<String>();
        listOfString.add("because");
        listOfString.add("can");
        listOfString.add("do");
        listOfString.add("must");
        listOfString.add("we");
        listOfString.add("what");
        String[] w= new String[listOfString.size()];

        for(int i =0;i <listOfString.size();i++) {
            w[i] = listOfString.get(i);
        }
        // That is for java 8
        //Arrays.sort(w, (a, b)->Integer.compare(a.length(), b.length()));

        for (int i = 0; i < w.length; i++) {
                for(int j=i+1;j<w.length;j++) {
                    String tempi = w[i];
                    String tempj = w[j];

                    if(tempj.length()<tempi.length()) {
                        w[i] =w[j];
                        w[j]=tempi;
                    }
                }
        }

       // That is for printing the sorted array
        for (int i = 0; i < w.length; i++) {
            System.out.println(w[i]);
        }

Output.

do
we
can
must
what
because

Kasasira · Accepted Answer · 2021-11-29 10:47:27Z

1

My answer is kind of irrelevant to this question (I wanted to sort and a list not an array) but since the answers provided here helped me solve my error, am going to leave it here for those that are facing the issue had.

Collections.sort(vehicles, (VehiclePayload vp1, VehiclePayload vp2) 
    -> Integer.compare(vp2.getPlateNumber().length(), vp1.getPlateNumber().length()));

answered Nov 29, 2021 at 10:47

Kasasira

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Comments

shailendra1118 · Accepted Answer · 2022-07-03 05:09:37Z

1

If you prefer Streams in Java8 onwards then you could use -

String input = "This is a beautiful world";
String[] arr = Arrays.stream(input.split(" "))
                     .sorted(Comparator.comparingInt(String::length))
                     .toArray(String[]::new);
// prints output
System.out.println(Arrays.toString(arr));

The output is -

[a, is, This, world, beautiful]

answered Jul 3, 2022 at 5:09

shailendra1118

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Comments

Vikrant Kashyap · Accepted Answer · 2016-03-08 12:03:50Z

0

import java.util.*;

class SortStringArray
{
    public static void main (String[] args) throws java.lang.Exception
    {
        String S = "No one could disentangle correctly";
        String W[] = S.split(" ");
        Arrays.sort(W, new StringLengthComparator());
        for(String str: W)
        System.out.println(str); //print Your Expected Result.
    }
}
 class StringLengthComparator implements Comparator<String>{ //Custom Comparator class according to your need

    @Override
        public int compare(String str1, String str2) {
            return str1.length() - str2.length();// compare length of Strings
        }
 }

answered Mar 8, 2016 at 12:03

Vikrant Kashyap

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Comments

Arif Shaik · Accepted Answer · 2022-06-28 12:49:54Z

0

//possible but not recommended;

 String sp[]="No one could disentangle correctly".split(" ");
    
    for(int i=0;i<sp.length-1;i++)
    {
        for(int j=i+1;j<sp.length;j++)
        {
            if(sp[i].length()>sp[j].length())
            {
                String temp=sp[i];
                sp[i]=sp[j];
                sp[j]=temp;
            }
        }
    }
    for(String i:sp) //will print content of array sp
        System.out.println(i);

answered Jun 28, 2022 at 12:49

Arif Shaik

1

Comments

Baba Fakruddin · Accepted Answer · 2022-09-07 18:03:52Z

0

ArrayList<String> list=new ArrayList<>();
    list.add("diamond");
    list.add("silver");
    list.add("gold");
    
    System.out.println(list.stream().sorted(Comparator.comparingInt(String::length)).collect(Collectors.toList()));

output:

[gold, silver, diamond]

answered Sep 7, 2022 at 18:03

Baba Fakruddin

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