对元素仍在最后位置的列表进行排序

Question

I have list of objects with data. I want the the particular object alone to stay last in the list when I am doing sorting. But its coming at the very first now.

Present:

Archived
Application
Run - time etc
Embedded Components IDE -
Debug, Compile and Build Tools
Initialization/Boot/Device Driver

Required

Application
Run - time etc
Embedded Components IDE -
Debug, Compile and Build Tools
Initialization/Boot/Device Driver
Archived

Collections.sort(existingList,new Comparator<Map<String, Object>>() {
  public int compare(Map<String, Object> o1,Map<String, Object> o2) {
    String archiveCheck = (String)o1.get("name");
    if(archiveCheck.equalsIgnoreCase("Archived"))
        return -10000;
    else
        return ((String) o1.get("name")).compareTo((String) o2.get("name"));
  }
});

Answer 1

Just return positive value whenever o1 is Archived

Collections.sort(existingList ,new Comparator<Map<String, Object>>() {
        public int compare(Map<String, Object> o1,Map<String, Object> o2) {
           String name1 = (String)o1.get("name");
           String name2 = (String)o2.get("name");

           if(name1.equalsIgnoreCase("Archived"))
             return 1;
           if(name2.equalsIgnoreCase("Archived"))
               return -1;
           return name1.compareTo(name2);
        }
    });

Edit We have to check the Archived value for both arguments in the Comparator. Updated the answer. So, in case of second argument matching Archived, then return negative value

Answer 2

A simple solution could be to remove the element you want to keep last from your list (or not add it in the first place), sort the list with its default comparator, then add the element at the end.

Answer 3

The most simply thing, but not such as effective as you probably need, which occurred to me is the idea about

• to iterate over original list - "Archived" copy to some temp , all others copy to newList ..
• Sort newList , and for the last insert "Archived" object (temp)