Sorting an array of String with custom ordering

Question

I have a String array:

 String[] str = {"ab" , "fog", "dog", "car", "bed"};
 Arrays.sort(str);
 System.out.println(Arrays.toString(str));

If I use Arrays.sort, the output is:

 [ab, bed, car, dog, fog]

But I need to implement the following ordering:

FCBWHJLOAQUXMPVINTKGZERDYS

I think I need to implement Comparator and override compare method:

 Arrays.sort(str, new Comparator<String>() {

        @Override
        public int compare(String o1, String o2) {
            // TODO Auto-generated method stub
            return 0;
        }
    });

How should I go about solving this?

Firstly, you should consider using a character array if your items are always one character in length. Using strings forces more error checking. — Duncan Jones
– Duncan Jones, Commented May 14, 2013 at 10:59
@DuncanJones No they are not, this was just an example. I changed them. Thanks — hamid
– hamid, Commented May 14, 2013 at 11:02
@Sam: so each individual character in the string needs to be sorted by the order and then "normal" string rules apply? How is "AA" sorted relative to "A"? Or "FC" relative to "FB"? — Joachim Sauer
– Joachim Sauer, Commented May 14, 2013 at 11:05
@JoachimSauer I don't know if I got your point correctly. But the order should be: "FC","FB","A","AA" — hamid
– hamid, Commented May 14, 2013 at 11:10

Egis · Accepted Answer · 2014-12-23 16:00:27Z

44

final String ORDER= "FCBWHJLOAQUXMPVINTKGZERDYS";

Arrays.sort(str, new Comparator<String>() {

    @Override
    public int compare(String o1, String o2) {
       return ORDER.indexOf(o1) -  ORDER.indexOf(o2) ;
    }
});

You can also add:

o1.toUpperCase()

If your array is case in-sensitive.

Apparently the OP wants to compare not only letters but strings of letters, so it's a bit more complicated:

    public int compare(String o1, String o2) {
       int pos1 = 0;
       int pos2 = 0;
       for (int i = 0; i < Math.min(o1.length(), o2.length()) && pos1 == pos2; i++) {
          pos1 = ORDER.indexOf(o1.charAt(i));
          pos2 = ORDER.indexOf(o2.charAt(i));
       }

       if (pos1 == pos2 && o1.length() != o2.length()) {
           return o1.length() - o2.length();
       }

       return pos1  - pos2  ;
    }

edited Dec 23, 2014 at 16:00

Egis

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answered May 14, 2013 at 10:57

Majid Laissi

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8 Comments

Joachim Sauer Over a year ago

careful: upper/lower-case confusion abound.

alex Over a year ago

@MajidL Should that be case in-sensitive?

hamid Over a year ago

@MajidL it doesn't seem to give correct result for: {"AA","FB","A","FC"}, it gives [AA, FB, FC, A].

Avi Over a year ago

It's not correct. The answer provided would work only if you gave it 1 length string or strings that are a substring of ORDER (for example - JLO)

maba Over a year ago

You could return o1.length() - o2.length() if pos1 - pos2 == 0 to have the shortest word first if they match like A and AA for example.

|

Avi · Accepted Answer · 2013-05-14 11:38:20Z

I would do something like this:

Put the letters in a HashTable (let's call it orderMap). Key is the letter, value is the index in ORDER.

And then:

Arrays.sort(str, new Comparator<String>() {

    @Override
    public int compare(String o1, String o2) {
        int length = o1.length > o2.length ? o1.length: o2.length
        for(int i = 0; i < length; ++i) {
           int firstLetterIndex = orderMap.get(o1.charAt(i));
           int secondLetterIndex = orderMap.get(o2.charAt(i));

           if(firstLetterIndex == secondLetterIndex) continue;

           // First string has lower index letter (for example F) and the second has higher index letter (for example B) - that means that the first string comes before
           if(firstLetterIndex < secondLetterIndex) return 1;
           else return -1;
        }

        return 0;
    }
});

For making it case-insensitive just do toUpperCase() to both strings at the beginning.

Community · Accepted Answer · 2017-05-23 12:02:03Z

0

Here you can find usefull link:

Using comparator to make custom sort

In your example instead comparing specific attributes of class you nedd to check possition of char in benchmarked String and base on this check if it's greather/equal/smaller.

edited May 23, 2017 at 12:02

CommunityBot

11 silver badge

answered May 14, 2013 at 11:03

Jacek

877 bronze badges

Comments

Mr.Nestorjava · Accepted Answer · 2017-09-01 07:13:55Z

Took my time to improve on the selected answer. This is more efficient

public static void customSort(final String order,String[] array){
String[] alphabets={"A","B","C","D","E","F","G","H","I","J","K","L","M","N","O","P","Q","R","S","T","U","V","W","X","Y","Z","0","1","2","3","4","5","6","7","8","9"};
    String keyword=order;
    for(int g=0; g<alphabets.length; g++){
    String one=alphabets[g];
    if(!keyword.toUpperCase().contains(one)){keyword=keyword+one;}
    }

final String finalKeyword=keyword;
Arrays.sort(array, new Comparator<String>() {

    @Override
   public int compare(String o1, String o2) {
       int pos1 = 0;
       int pos2 = 0;
       for (int i = 0; i < Math.min(o1.length(), o2.length()) && pos1 == pos2; i++) {
          pos1 = finalKeyword.toUpperCase().indexOf(o1.toUpperCase().charAt(i));
          pos2 = finalKeyword.toUpperCase().indexOf(o2.toUpperCase().charAt(i));
       }

       if (pos1 == pos2 && o1.length() != o2.length()) {
           return o1.length() - o2.length();
       }

       return pos1  - pos2  ;
    }
});
//Arrays.sort(array, Collections.reverseOrder());
}

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