Sort a Single String in Java

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.

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"
  }
}

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());

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"

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);
}

    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.

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);

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