Java – Sort a single String in Java

javasortingstring

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

String s = "edcba"  ->  "abcde"

Best Solution

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.

This is the style that Microsoft tends to use in their examples.

It appears that the guidance in this area may have changed, as StyleCop now enforces the use of the C# specific aliases.

Java – Is Java “pass-by-reference” or “pass-by-value”

Java is always pass-by-value. Unfortunately, when we deal with objects we are really dealing with object-handles called references which are passed-by-value as well. This terminology and semantics easily confuse many beginners.

It goes like this:

public static void main(String[] args) {
    Dog aDog = new Dog("Max");
    Dog oldDog = aDog;

    // we pass the object to foo
    foo(aDog);
    // aDog variable is still pointing to the "Max" dog when foo(...) returns
    aDog.getName().equals("Max"); // true
    aDog.getName().equals("Fifi"); // false
    aDog == oldDog; // true
}

public static void foo(Dog d) {
    d.getName().equals("Max"); // true
    // change d inside of foo() to point to a new Dog instance "Fifi"
    d = new Dog("Fifi");
    d.getName().equals("Fifi"); // true
}

In the example above aDog.getName() will still return "Max". The value aDog within main is not changed in the function foo with the Dog "Fifi" as the object reference is passed by value. If it were passed by reference, then the aDog.getName() in main would return "Fifi" after the call to foo.

Likewise:

public static void main(String[] args) {
    Dog aDog = new Dog("Max");
    Dog oldDog = aDog;

    foo(aDog);
    // when foo(...) returns, the name of the dog has been changed to "Fifi"
    aDog.getName().equals("Fifi"); // true
    // but it is still the same dog:
    aDog == oldDog; // true
}

public static void foo(Dog d) {
    d.getName().equals("Max"); // true
    // this changes the name of d to be "Fifi"
    d.setName("Fifi");
}

In the above example, Fifi is the dog's name after call to foo(aDog) because the object's name was set inside of foo(...). Any operations that foo performs on d are such that, for all practical purposes, they are performed on aDog, but it is not possible to change the value of the variable aDog itself.

For more information on pass by reference and pass by value, consult the following SO answer: https://stackoverflow.com/a/430958/6005228. This explains more thoroughly the semantics and history behind the two and also explains why Java and many other modern languages appear to do both in certain cases.

Best Solution

Related Solutions

C# – the difference between String and string in C#

This is the style that Microsoft tends to use in their examples.

Java – Is Java “pass-by-reference” or “pass-by-value”

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