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Sorting a list with the element still in first position

I have a list of strings:

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

I'm doing some processing on a list, and I want to sort that list, but I want "general" to always be in the first position. thanks;)

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4 answers


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



Collections.sort(list, new Comparator<String>()
  {
     int compare(String o1, String o2)
     {
         if (o1.equals(o2)) // update to make it stable
           return 0;
         if (o1.equals("general"))
           return -1;
         if (o2.equals("general"))
           return 1;
         return o1.compareTo(o2);
     }
});
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Do Collections.sort

instead subList

.

    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 references

  • subList(int fromIndex, int toIndex)

    • Returns a representation of the portion of this list between the specified fromIndex

      , inclusive and toIndex

      exclusive. The returned list is maintained by this list, so non-structural changes to the returned list are reflected in this list and vice versa. The returned list supports all additional list operations supported by this list.


If the special item is not at index 0, just put it there before sorting it like this:

    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 references

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

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You can use the following code snippet, but it may have performance / memory issues for very large 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);
}
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