Comparing two string and sorting them in alphabetical order

Question

I want to compare two strings and sort them in alphabetical order. I am currently creating two arrays with the strings and sorting one them comparing the two arrays.

String a="LetterA";
String b="ALetterB";
String[] array1={a.toLowerCase(),b.toLowerCase()};
String[] array2={a.toLowerCase(),b.toLowerCase()};
Arrays.sort(array2);
if (Arrays.equals(array1, array2)){
    System.out.println(a+" is before "+b);
}
else{
    System.out.println(b+" is before "+a);
}

This works but it is time and memory consuming. I would appreciate if anyone can suggest a better way to do this.

what is the goal? Could you elaborate more ? – Amit Sharma Dec 07 '13 at 20:00 — Amit Sharma, Dec 07 '13 at 20:00

Amit Sharma · Accepted Answer · 2013-12-07T20:08:27.710

23

Hint: All basic data type classes in java implement Comparable interface.

String a="LetterA";
String b="ALetterB";
int compare = a.compareTo(b);
if (compare < 0){
    System.out.println(a+" is before "+b);
}
else if (compare > 0) {
    System.out.println(b+" is before "+a);
}
else {
    System.out.println(b+" is same as "+a);
}

edited Dec 07 '13 at 20:08

answered Dec 07 '13 at 20:02

Amit Sharma

5,844
5
25
34

2

If you are not in plain ascii mode this strategy wont wokr right. See http://stackoverflow.com/a/12927962/2087666 – Remi Morin Dec 19 '14 at 16:35
Caution: ASCII values of capital letters are less than that of small letters. If cases are like 1> a="Ax" and b="aa" or 2> a="aa" and b="AA"...The results will be a contradiction to expected alphabetical sorting. Better convert both strings in a common "CASE" and then compare. – Deepeshkumar Sep 19 '15 at 11:55

score 2 · Answer 2 · answered Dec 07 '13 at 20:02

2

int compare = a.compareTo(b);
if (compare < 0){
    System.out.println(a + " is before " +b);
} else if (compare > 0) {
    System.out.println(b + " is before " +a);
} else {
    System.out.println("Strings are equal")
}

answered Dec 07 '13 at 20:02

irla

479
3
13

add a case of equality too :) – Amit Sharma Dec 07 '13 at 20:02
Done - also cached comparison. – irla Dec 07 '13 at 20:04

score 1 · Answer 3 · answered Jan 02 '17 at 16:26

1

If you just look for an easy and elegant code and you do not want to preoptimize, in java 8 you can do like this:

String[] sorted = Stream.of(a, b).sorted().toArray(String[]::new);
System.out.println(sorted[0] + " is before " + sorted[1]);

answered Jan 02 '17 at 16:26

Anton Ivinskyi

402
5
17

score 0 · Answer 4 · edited Jul 28 '18 at 09:25

0

Here is the code for the problem.

 public static void main(String[] args) {
    String a="Angram";
    String b="Angram";

    if(a.length()==b.length()) {
        char c[]=a.toCharArray();
        char d[]=b.toCharArray();
        Arrays.sort(c);
        Arrays.sort(d);
        a=new String(c);
        b=new String(d);
        if(a.equals(b)) 
            System.out.print("Yes");
        else
            System.out.print("No");
    }
    else {
        System.out.print("No");
    }
}

edited Jul 28 '18 at 09:25

Avishek Bhattacharya

6,534
3
34
53

answered Jul 28 '18 at 06:48

Siva Sankaran

1

Please explain the algorithm. – Avishek Bhattacharya Jul 28 '18 at 06:55

Comparing two string and sorting them in alphabetical order

4 Answers4