Sort a list with element still in first position

Question

I have a String list:

List<String> listString  = new ArrayList<String>();
listString.add("faq");
listString.add("general");
listString.add("contact");

I do some processing on the list and I want to sort this list but I want "general" to always end up in first position. Thx ;)

Answer 1

Sort the list without having "general" inside it and then add it to the beginning.

Answer 2

Do Collections.sort on its subList instead.

    List<String> list = new ArrayList<String>(
        Arrays.asList("Zzz...", "Two", "One", "Three")
    );
    Collections.sort(list.subList(1, list.size()));
    System.out.println(list);
    // "[Zzz..., One, Three, Two]"

API links

• subList(int fromIndex, int toIndex)
  • Returns a view of the portion of this list between the specified fromIndex, inclusive, and toIndex, exclusive. The returned list is backed by this list, so non-structural changes in the returned list are reflected in this list, and vice-versa. The returned list supports all of the optional list operations supported by this list.

---

If the special element is not at index 0, then simply put it there before you sort it as follows:

    List<String> list = new ArrayList<String>(
        Arrays.asList("Four", "Five", "Zzz...", "Two", "One", "Three")
    );
    Collections.swap(list, list.indexOf("Zzz..."), 0);
    Collections.sort(list.subList(1, list.size()));
    System.out.println(list);
    // "[Zzz..., Five, Four, One, Three, Two]"

API links

• Collections.swap(List<?> list, int i, int j)
  • Swaps the elements at the specified positions in the specified list.

Answer 3

I like @Petar's approach, but another approach would be to sort it using a custom Comparator that always said that "general" was before whatever it was being compared to.

Collections.sort(list, new Comparator<String>()
  {
     int compare(String o1, String o2)
     {
         if (o1.equals("general"))
           return -1;
         if (o2.equals("general"))
           return 1;
         return o1.compareTo(o2);
     }
});

Answer 4

You can use the following code snippet but it might have perfomance/memory problems for the very big lists.

public static List<String> sortSpecial(List<String> list, final String alwaysOnTopItem) {
    list.remove(alwaysOnTopItem);
    Collections.sort(list);

    List<String> result = new ArrayList<String>(list.size() + 1);
    result.add(alwaysOnTopItem);
    result.addAll(list);

    return result;
}

public static void main(String[] args) {
    List<String> listString = new ArrayList<String>();
    listString.add("faq");
    listString.add("general");
    listString.add("contact");
    String alwaysOnTopItem = "general";
    List<String> sortedList = sortSpecial(listString, alwaysOnTopItem);
    System.out.println(sortedList);
}