سؤال

I have an ArrayList in Java:

{"deleteItem", "createitem", "exportitem", "deleteItems", "createItems"}

I want to move all string which contains delete to the end of the list, so I would get the next:

{"createitem", "exportitem", "createItems", "deleteItem", "deleteItems"}`

I can create two sublists - one for the words which contain the 'delete' word, and one for the others, and then merge them, but I search for a more efficient way.

هل كانت مفيدة؟

المحلول

Use custom Comparator:

List<String> strings = Arrays.asList(
        "deleteItem", "createitem", "exportitem", "deleteItems", "createItems"
        );
Comparator<String> comparator = new Comparator<String>() {
    @Override
    public int compare(final String o1, final String o2) {
        if (o1.contains("delete") && !o2.contains("delete")) {
            return 1;
        }else if (!o1.contains("delete") && o2.contains("delete")) {
            return -1;
        }
        return 0;
    }
};
Collections.sort(strings, comparator);
System.out.println(strings);

نصائح أخرى

If you want something efficient and need to remove elements in the beginning and middle of a List I would suggest using a LinkedList instead of a array list. That would avoid rewriting the underlying array for each remove operation.

Then, you simply iterate on the list, calling remove and addLast for any string that contains delete.

Of course, this is only OK if there is nothing preventing you from replacing your ArrayList with a LinkedList.

You only want to put the elements with delete at the end of the list so ordering is O(nlogn) while we could do this in one pass, in O(n) (although using a new list). We could create a new LinkedLIst and pass through the original list adding the elements with "delete" at the end and the others at the begginging.

    LinkedList<String> orderedList = new LinkedList<>();
    for(String e:originalList){
        if (e.indexOf("delete")>=0) {
            orderedList.addLast(e);
        } else {
            orderedList.addFirst(e);
        }
    }

This is faster than sorting.

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