Is there a native way to sort a String by its contents in java? E.g. 

String s = "edcba" -> "abcde"

13 Answers

toCharArray followed by Arrays.sort followed by a String constructor call:

import java.util.Arrays; public class Test { public static void main(String[] args) { String original = "edcba"; char[] chars = original.toCharArray(); Arrays.sort(chars); String sorted = new String(chars); System.out.println(sorted); } }

EDIT: As tackline points out, this will fail if the string contains surrogate pairs or indeed composite characters (accent + e as separate chars) etc. At that point it gets a lot harder... hopefully you don't need this :) In addition, this is just ordering by ordinal, without taking capitalisation, accents or anything else into account.

8

No there is no built-in String method. You can convert it to a char array, sort it using Arrays.sort and convert that back into a String.

String test= "edcba"; char[] ar = test.toCharArray(); Arrays.sort(ar); String sorted = String.valueOf(ar);

Or, when you want to deal correctly with locale-specific stuff like uppercase and accented characters:

import java.text.Collator; import java.util.Arrays; import java.util.Comparator; import java.util.Locale; public class Test { public static void main(String[] args) { Collator collator = Collator.getInstance(new Locale("fr", "FR")); String original = "éDedCBcbAàa"; String[] split = original.split(""); Arrays.sort(split, collator); String sorted = ""; for (int i = 0; i < split.length; i++) { sorted += split[i]; } System.out.println(sorted); // "aAàbBcCdDeé" } }
6

In Java 8 it can be done with:

String s = "edcba".chars() .sorted() .collect(StringBuilder::new, StringBuilder::appendCodePoint, StringBuilder::append) .toString();

A slightly shorter alternative that works with a Stream of Strings of length one (each character in the unsorted String is converted into a String in the Stream) is:

String sorted = Stream.of("edcba".split("")) .sorted() .collect(Collectors.joining());
2

Convert to array of chars → Sort → Convert back to String:

String s = "edcba"; char[] c = s.toCharArray(); // convert to array of chars java.util.Arrays.sort(c); // sort String newString = new String(c); // convert back to String System.out.println(newString); // "abcde"
3

A more raw approach without using sort Arrays.sort method. This is using insertion sort.

public static void main(String[] args){ String wordSt="watch"; char[] word=wordSt.toCharArray(); for(int i=0;i<(word.length-1);i++){ for(int j=i+1;j>0;j--){ if(word[j]<word[j-1]){ char temp=word[j-1]; word[j-1]=word[j]; word[j]=temp; } } } wordSt=String.valueOf(word); System.out.println(wordSt); }
5
 String a ="dgfa"; char [] c = a.toCharArray(); Arrays.sort(c); return new String(c);

Note that this will not work as expected if it is a mixed case String (It'll put uppercase before lowercase). You can pass a comparator to the Sort method to change that.

1

Procedure :

  1. At first convert the string to char array
  2. Then sort the array of character
  3. Convert the character array to string
  4. Print the string

Code snippet:

 String input = "world"; char[] arr = input.toCharArray(); Arrays.sort(arr); String sorted = new String(arr); System.out.println(sorted);
2

Question: sort a string in java

public class SortAStringInJava { public static void main(String[] args) { String str = "Protijayi"; // Method 1 str = str.chars() // IntStream .sorted().collect(StringBuilder::new, StringBuilder::appendCodePoint, StringBuilder::append).toString(); System.out.println(str); // Method 2 str = Stream.of(str.split(" ")).sorted().collect(Collectors.joining()); System.out.println(str); } } 
str.chars().boxed().map(Character::toString).sorted().collect(Collectors.joining())

or

s.chars().mapToObj(Character::toString).sorted().collect(Collectors.joining())

or

Arrays.stream(str.split("")).sorted().collect(Collectors.joining())

A solution that uses the Stream API and also handles Unicode supplementary characters:

public static String sort(final String s) { return s.codePoints() .sorted() .collect(StringBuilder::new, StringBuilder::appendCodePoint, StringBuilder::append) .toString(); }

You can also write up a counting sort algorithm to sort all the characters in an array if you would like to reduce your worst-case time complexity from nlogn to n

public static void main(String[] args) { String str = "helloword"; char[] arr; List<Character> l = new ArrayList<Character>(); for (int i = 0; i < str.length(); i++) { arr = str.toCharArray(); l.add(arr[i]); } Collections.sort(l); str = l.toString(); System.out.println(str); str = str.replaceAll("\\[", "").replaceAll("\\]", "") .replaceAll("[,]", ""); System.out.println(str); }

Without using Collections in Java:

import java.util.Scanner; public class SortingaString { public static String Sort(String s1) { char ch[]=s1.toCharArray(); String res=" "; for(int i=0; i<ch.length ; i++) { for(int j=i+1;j<ch.length; j++) { if(ch[i]>=ch[j]) { char m=ch[i]; ch[i]=ch[j]; ch[j]=m; } } res=res+ch[i]; } return res; } public static void main(String[] args) { Scanner sc=new Scanner(System.in); System.out.println("enter the string"); String s1=sc.next(); String ans=Sort( s1); System.out.println("after sorting=="+ans); } }

Output:

enter the string==

sorting

after sorting== ginorst

1