Why doesn't my selection sort display a properly ordered array?

Question

I have to write a selection sort. This is what I came up with

     public static void sortString(String[] stringArray) {
    int max = DataManager.getArraySizeExcludingNull(stringArray), smallest = 0;
    String temp;
    for (int i = 0; i < max; i++) {

        smallest = i;

        for (int j = 0; j < max; j++) {

            if (stringArray[j].compareTo(stringArray[smallest]) < 0) {

                smallest = j;
            }

        }

        temp = stringArray[i];
        stringArray[i] = stringArray[smallest];
        stringArray[smallest] = temp;

    }
}

I have a list with 45 names (I've checked that max is of size 45 and that each index is a string). The method seems to separate the array in half, in a lexicographically correct order in 2 parts? This is my output.

 Kate

 Leroy

 Nicola

 Nancy

 Oprah

 Peter

 Quinton

 Richard

 Kelly

 Sven

 Theo

 Terry

 Violet

 Wilma

 Zed

 Allan

 Bob

 Steve

 Nigel

 Neil

 Fred

 Sally

 Glenn

 Gary

 Heather

 Horatio

 Ivan

 Ingrid

 Susy

 Lindsay

 Mitch

 Shelly

 David

 Kirsten

 Sarah

 Janet

 Barbra

 Carrie

 Jacob

 Elenor

 Evan

 John

 Mike

 Josh

Aaron

Answer 1

You need to loop over the rest of the data to find the minimum, not all the data, i.e. j should start from i+1, not 0, otherwise you just end up putting the minimum into the first position, then moving it to the second, then the third and so on, leaving the array completely unsorted, except that the minimum will be at the end.

And using DataManager.getArraySizeExcludingNull(stringArray) is probably wrong for this problem (judging by the name of the function), as you're treating max as the last index, not the number of non-null values. You need to simply use stringArray.length and/or change your function appropriately.