比较器未给出预期结果

Question

Below is the code i wrote for placing all anagrams next to each other, in a collection of strings. The output is not sorted as expected.In fact, the output is the same as the input. Where am I going wrong ?

package set2;

import java.util.Arrays;
import java.util.Comparator;

public class printAllAnagrams {
    public static void main(String[] args) {
        String[] s = { "Harsha", "ant", "sha", "tna", "ash" };
        sortAnagrams(s);
        for (String e : s) {
            System.out.println(e);
        }
    }

    private static void sortAnagrams(String[] s) {
        Arrays.sort(s, new Comparator<String>() {

            @Override
            public int compare(String s1, String s2) {
                s1.toLowerCase();
                s2.toLowerCase();

                if (s1.length() != s2.length()) {
                    return -1;
                } else {
                    char[] s1_char = s1.toCharArray();
                    char[] s2_char = s2.toCharArray();
                    Arrays.sort(s1_char);
                    Arrays.sort(s2_char);

                    for (int i = 0; i < s1_char.length; i++) {
                        if (s1_char[i] != s2_char[i]) {
                            return -1;
                        }
                    }
                }
                return 0;
            }

        });

    }
}

Answer 1

Without looking too closely, this

 s1.toLowerCase();

doesn't alter s1 but rather returns a new string which is a lower case variant of s1 . In Java Strings are immutable . So you need to collect and work with the value returned from the above.

Answer 2

This can't be right for sure:

if (s1.length() != s2.length()) {
    return -1;
}

It would mean that if s1.length() != s2.length() s1 < s2 and s2 < s1 .

What I think you mean to do is:

public int compare(String s1, String s2) {
    if (s1.length() == s2.length()) {                   
        char[] s1_char = s1.toLowerCase().toCharArray();
        char[] s2_char = s2.toLowerCase().toCharArray();
        Arrays.sort(s1_char);
        Arrays.sort(s2_char);

        for (int i = 0; i < s1_char.length; i++) {
            if (s1_char[i] != s2_char[i]) {
                return (int)(s1_char[i] - s2_char[i]);
            }
        }
        return 0;
    } else {
        return s1.length() - s2.length();
    }
}

Answer 3

Why don't you just do the following:

        @Override
        public int compare(String s1, String s2) {
            return s1.toLowerCase().compareTo(s2.toLowerCase());
        }

Answer 4

Your comparator is not stable at all.

First of all you return -1 if the lengths are different. This means that depending on the operand order you might find that "asbd" > "ash" or the opposite.

Also you do the same with the char comparison.

if (s1_char[i] != s2_char[i]) {
   return -1;
}

Substitute this with:

if (s1_char[i] != s2_char[i]) {
  return s1_char[i] > s2_char[i] ? 1 : -1;
}

Use the same pattern for the length comparison.

EDIT Returning -1 from a compare methods, means that you find the first operand smaller than the second as per the documentation .

Answer 5

Strings are immutable so just calling the method on a String object doesn't change the String itself. You should use the following:

s1 = s1.toLowerCase();
s2 = s2.toLowerCase();

Answer 6

From the javadoc :

The implementor must ensure that sgn(compare(x,y)) == -sgn(compare(y,x)) for all x and y . (This implies that compare(x,y) must throw an exception if and only if compare(y,x) throws an exception.).

This basically means if you call compare(s1,s2) it has to yield -1 * compare(s2,s1) and neither of your return -1; statements follow this. Instead of that, you can compare the ints and characters to each other and return that value, for instance this code works (instead of the return -1; ):

return Integer.compare(s1.length(),s2.length()); //for the ints

return Character.compare(s1_char[i],s2_char[i]); //for the chars

Also take a look at Vishal's answer, that's another bug.