排序ArrayList-IndexOutOfBoundsException -Java

Question

I'm trying to sort an ArrayList with strings(PlayersNames) and imageIcons(PlayersIcons) based on the values i store in an other arrayList with integers(results). As you can see i get an indexOutOfBoundsException but i cant understand why. Maybe the earling of the morning makes me not to see plain things.

ArrayList<String> PlayersNames=new ArrayList<String>;
ArrayList<ImageIcon> PlayersIcons=new ArrayList<ImageIcons>;

    public void sortPlayers(ArrayList<Integer> results){
        String tmp;
        ImageIcon tmp2;
        for (int i=0; i<PlayersNames.size(); i++) {
            for (int j=PlayersNames.size(); j>i; j--) {

                if (results.get(i) < results.get(i+1) ) {       //IndexOutOfBoundsException!

                    tmp=PlayersNames.get(i+1);
                    PlayersNames.set(i+1,PlayersNames.get(i));
                    PlayersNames.set(i,tmp);

                    tmp2=PlayersIcons.get(i+1);
                    PlayersIcons.set(i+1,PlayersIcons.get(i));
                    PlayersIcons.set(i,tmp2);
                }
            }
        }
    }

Answer 1

When the loop gets to the end of the arrayList, you are trying to get an item past the end of the list. On this line:

if (results.get(i) < results.get(i+1) ) {

If i=9, with an arrayList with 10 items, results.get(9) will give you the last item in the list. results.get(10) will try to get something that doesn't exist.

Answer 2

You can use Collections.sort(Pass ArrayList Here) , You need not to write your own method. Java provides it.

Answer 3

Eventually i can hold a value of PlayersNames.size()-1 . I can only assume that results has the same size as PlayersNames , or rather PlayersNames.size() == results.size() .

If this is true, then eventually you're asking for the results.size() -th element (by doing results.get(i+1) ) in results , which is one more than results holds, thus an IndexOutOfBoundsException is being thrown.

Said more concisely, if results holds N items, then the N -th item is accessed using the index N-1 , but you're trying to access the item at index N , which doesn't exist.

Try changing your outer loop to:

for (int i=0; i<PlayersNames.size()-1; i++) {

to prevent the overrun.

Also, your inner loop appears to be unused, but if you try to access something in one of your arrays using the first value of j you're likely to run into the same issue for the same reason.

Answer 4

The very last iteration of the for loop, i will equal PlayersNames.size() - 1 . On the line where the error occurred, you're calling results.get(i + 1) , which evaluates to results.get(PlayersNames.size()) .

Answer 5

Couple of mistakes that I can see :

1)

i<PlayersNames.size()

that is good, but then you use

i+1

EVERYWHERE (not only in the if), so when you reach the last element you will always hit an indexOutOfBoundsException.

Either reduce the range of i or remove the +1;

2) you declare a variable

j

that you never use...

Answer 6

Many people have given the correct reasons.

There are many ways to correct this program. Easiest one is to iterate outer loop only till n-1 (where n is the size of the ArrayList)

    for (int i=0; i<PlayersNames.size()-1; i++) {

Answer 7

Using Map like:

Map <String, ImageIcon>

may be more useful for sorting, rather than using two ArrayLists.