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"
  }
}

    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.

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