对元素仍处于第一个位置的列表进行排序

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

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(o2)) // update to make it stable
           return 0;
         if (o1.equals("general"))
           return -1;
         if (o2.equals("general"))
           return 1;
         return o1.compareTo(o2);
     }
});

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

对列表进行排序而不在其中包含“常规”，然后将其添加到开头。

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);
}

Answer 5

Nice! Question. If we want to fix the positions for the first and last elements then we can enhance the @Paul Tomblin's approach as like below one. Please copy the code and try it.

import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;
import java.util.List;

public class CustomComparatorToFixedPosition {

  public static void main(String[] args) {
    List<Emp> empNames = new ArrayList<>();
    empNames.add(new Emp("name1"));
    empNames.add(new Emp("LastName"));
    empNames.add(new Emp("name2"));
    empNames.add(new Emp("FirstName"));
    empNames.add(new Emp("name3"));
    getSortedName(empNames).stream().forEach(emp -> System.out.println(emp.getName()));
  }

  public static List<Emp> getSortedName(List<Emp> empNames) {
    String first = "FirstName";
    String last = "LastName";
    Collections.sort(empNames, new Comparator<Emp>() {
      public int compare(Emp o1, Emp o2) {
        if (o1.getName().equalsIgnoreCase(first) || o2.getName().equalsIgnoreCase(last))
          return -1;
        if (o2.getName().equalsIgnoreCase(first) || o1.getName().equalsIgnoreCase(last))
          return 1;
        return o1.getName().compareTo(o2.getName());
      }
    });
    return empNames;
  }
}

class Emp {

  private String name;

  Emp(String name) {
    this.name = name;
  }

  public String getName() {
    return name;
  }

  public void setName(String name) {
    this.name = name;
  }

}