Using a selection sort on an ArrayList of Strings

Question

I have a program that is counting the frequencies of words taken from a txt file and they are being stored in an ArrayList. I am extremely unfamiliar with using a selection sort but it is the type of sort that I am being asked to use. I have looked at multiple selection sorts, but mine is falling short somewhere along the line.

This is my actual sort.

private void sort() {

    for (int i = 0; i < wordArray.size() - 1; i++) {
        for (int j = i + 1; j < wordArray.size(); j++) {
            if (wordArray.get(i).compareTo(wordArray.get(j)) == 1) {

                Word temp = wordArray.get(i);
                wordArray.set(i, wordArray.get(j));
                wordArray.set(j, temp);
            }
        }
    }
}

This is my comparison of strings (I am pretty sure the logic error is in here).

public int compareTo(Word w) {

    for (int i = 0; i < this.word.length() - 1; i++) {
        if (i <= w.word.length() - 1) {
            if (this.word.charAt(i) < w.word.charAt(i)) {
                return -1;
            } else if (this.word.charAt(i) > w.word.charAt(i)){
                return 1;
            }
        }
    }
    return -1;
}

Word is a class that has the String variable "word". Any tips would be greatly appreciated :)

What is going wrong with this? i.e. what is the incorrect behaviour? Have you tried stepping through your code in a debugger? — Oliver Charlesworth
– Oliver Charlesworth, Commented May 2, 2011 at 0:44
above, add, a, advanced, ago, all, as, altogether, and, any, are, be. This is a sample of how it is currently turning out. The document has about 138 unique words in it. — Alex
– Alex, Commented May 2, 2011 at 0:48
I don't think you want to implement the sort on your own. I also believe that the Word.compareTo can simply delegate to the Word's String.compareTo. — ditkin
– ditkin, Commented May 2, 2011 at 0:49
If you are suspicious that your compareTo method is broken, why not try testing it in isolation? — Oliver Charlesworth
– Oliver Charlesworth, Commented May 2, 2011 at 0:50
@ditkin That is why I am having a hard time finding useful information from Google, but I am explicitly told to manually use the selection sort. — Alex
– Alex, Commented May 2, 2011 at 0:51

corsiKa · Accepted Answer · 2011-05-02 00:46:28Z

2

Why not just use this

public int compareTo(Word w) {
    return this.word.compareTo(w.word);
}

answered May 2, 2011 at 0:46

corsiKa

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1 Comment

Alex Over a year ago

Actually after changing the compareTo AND my sort to if (wordArray.get(i).compareTo(wordArray.get(j)) >=1) It works :) Thank you!

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