List an Array of Strings in alphabetical order

Question

I have a program which has the user inputs a list of names. I have a switch case going to a function which I would like to have the names print off in alphabetical order.

public static void orderedGuests(String[] hotel)
{
  //??
}

I have tried both

Arrays.sort(hotel);
System.out.println(Arrays.toString(hotel));

and

java.util.Collections.sort(hotel);

Answer 1

Weird, your code seems to work for me:

import java.util.Arrays;

public class Test
{
    public static void main(String[] args)
    {
        // args is the list of guests
        Arrays.sort(args);
        for(int i = 0; i < args.length; i++)
            System.out.println(args[i]);
    }
}

I ran that code using "java Test Bobby Joe Angel" and here is the output:

$ java Test Bobby Joe Angel
Angel
Bobby
Joe

Answer 2

The first thing you tried seems to work fine. Here is an example program.
Press the "Start" button at the top of this page to run it to see the output yourself.

import java.util.Arrays;

public class Foo{
    public static void main(String[] args) {
        String [] stringArray = {"ab", "aB", "c", "0", "2", "1Ad", "a10"};
        orderedGuests(stringArray);
    }

    public static void orderedGuests(String[] hotel)
    {
        Arrays.sort(hotel);
        System.out.println(Arrays.toString(hotel));
    }
}

Answer 3

You can just use Arrays#sort() , it's working perfectly. See this example :

String [] a = {"English","German","Italian","Korean","Blablablabla.."};
//before sort
for(int i = 0;i<a.length;i++)
{
  System.out.println(a[i]);
}
Arrays.sort(a);
System.out.println("After sort :");
for(int i = 0;i<a.length;i++)
{
  System.out.println(a[i]);
}

Answer 4

java.util.Collections.sort(listOfCountryNames, Collator.getInstance());

Answer 5

Here is code that works:

import java.util.Arrays;
import java.util.Collections;

public class Test
{
    public static void main(String[] args)
    {
        orderedGuests1(new String[] { "c", "a", "b" });
        orderedGuests2(new String[] { "c", "a", "b" });
    }

    public static void orderedGuests1(String[] hotel)
    {
        Arrays.sort(hotel);
        System.out.println(Arrays.toString(hotel));
    }

    public static void orderedGuests2(String[] hotel)
    {
        Collections.sort(Arrays.asList(hotel));
        System.out.println(Arrays.toString(hotel));
    }

}

Answer 6

By alphabetical-order I assume the order to be : A|a < B|b < C|c... Hope this is what @Nick is(or was) looking for and the answer follows the above assumption.

I would suggest to have a class implement compare method of Comparator-interface as :

public int compare(Object o1, Object o2) {
    return o1.toString().compareToIgnoreCase(o2.toString());
}

and from the calling method invoke the Arrays.sort method with custom Comparator as :

Arrays.sort(inputArray, customComparator);

Observed results: input Array : "Vani","Kali", "Mohan","Soni","kuldeep","Arun"

output(Alphabetical-order) is : Arun, Kali, kuldeep, Mohan, Soni, Vani

Output(Natural-order by executing Arrays.sort(inputArray) is : Arun, Kali, Mohan, Soni, Vani, kuldeep

Thus in case of natural ordering, [Vani < kuldeep] which to my understanding of alphabetical-order is not the thing desired.

for more understanding of natural and alphabetical/lexical order visit discussion here

Answer 7

You can use Arrays.sort() method. Here's the example,

import java.util.Arrays;

public class Test 
{
    public static void main(String[] args) 
    {
        String arrString[] = { "peter", "taylor", "brooke", "frederick", "cameron" };
        orderedGuests(arrString);
    }

    public static void orderedGuests(String[] hotel)
    {
        Arrays.sort(hotel);
        System.out.println(Arrays.toString(hotel));
    }
}

Output

[brooke, cameron, frederick, peter, taylor]

Answer 8

CompareTo() method: The two strings are compared based on Unicode character values.

import java.util.*;

public class Test {

int n,i,temp;
String names[n];

public static void main(String[] args) {
String names[5] = {"Brian","Joshua","Louis","David","Marcus"};

for(i=0;i<5;i++){
    for(j=i+1;i<n;j++){
        if(names[i].CompareTo(names[j]>0) {
             temp=names[i];
             names[i]=names[j];
             names[j]=temp;
         } else
             System.out.println("Alphabetically Ordered");                               
         }
     }                              
}

Answer 9

**//With the help of this code u not just sort the arrays in alphabetical order but also can take string from user or console or keyboard

import java.util.Scanner;
import java.util.Arrays;
public class ReadName
{
final static int ARRAY_ELEMENTS = 3;
public static void main(String[] args)
{
String[] theNames = new String[5];
Scanner keyboard = new Scanner(System.in);
System.out.println("Enter the names: ");
for (int i=0;i<theNames.length ;i++ )
{           
theNames[i] = keyboard.nextLine();
}
System.out.println("**********************");
Arrays.sort(theNames);
for (int i=0;i<theNames.length ;i++ )
{
System.out.println("Name are " + theNames[i]);
}
}
}**

Answer 10

Arrays.sort(stringArray); This sorts the string array based on the Unicode characters values. All strings that contain uppercase characters at the beginning of the string will be at the top of the sorted list alphabetically followed by all strings with lowercase characters. Hence if the array contains strings beginning with both uppercase characters and lowercase characters, the sorted array would not return a case insensitive order alphabetical list

String[] strArray = { "Carol", "bob", "Alice" };
Arrays.sort(strList);
System.out.println(Arrays.toString(hotel));

Output is : Alice, Carol, bob,

If you require the Strings to be sorted without regards to case, you'll need a second argument, a Comparator, for Arrays.sort(). Such a comparator has already been written for us and can be accessed as a static on the String class named CASE_INSENSITIVE_ORDER.

String[] strArray = { "Carol", "bob", "Alice" };
Arrays.sort(stringArray, String.CASE_INSENSITIVE_ORDER);
System.out.println(Arrays.toString(strArray ));

Output is : Alice, bob, Carol

Answer 11

 public static String[] textSort(String[] words) {
    for (int i = 0; i < words.length; i++) {
        for (int j = i + 1; j < words.length; j++) {
            if (words[i].compareTo(words[j]) > 0) {
                String temp = words[i];
                words[i] = words[j];
                words[j] = temp;
            }
        }
    }

    return words;
}