在Android中列出：重命名重复项

Question

I want to create lists as follows:

• list-item
• list-item-1
• list-item-2
• list-myitem
• list-myItem-1

As above with list-item , if the item's name is the same in next row, then it will add the suffix 1 .

like this done upto this enter image description here

Answer 1

OK, so you want to ask how to implement this. The most simple solution I can think of is using a Map<String, Int> to keep track of the item name. Your code will look somewhat like this:

final Map<String, Integer> listNameMap = new HashMap<String, Integer>();

When displaying a list item name, you'll do:

final String displayedName;
if (listNameMap.containsKey(name)) {
   Integer newValue = listNameMap.get(name).intValue() + 1;
   listNameMap.put(name, newValue);
   displayedName = name + "-" + String.valueOf(newValue);
} else {
   listNameMap.put(name, 0);
   displayedName = name;
}
// then use the displayedName to display

Answer 2

i have tried with sample data and below solution worked for me.But there can more efficient solution also. This solution doesn't require ordering of data.

List<String> parent =new  ArrayList<String>();

    parent.add("R");
    parent.add("I");
    parent.add("S");

    parent.add("R");
    parent.add("I");
    parent.add("S");
    parent.add("R");
    parent.add("I");
    parent.add("S");
    parent.add("R");
    parent.add("R");
    parent.add("I");

    HashSet<String> set = new HashSet<>(parent);
    ArrayList<String> valuesList = new ArrayList<>();
    valuesList .addAll(set);



    Log.d("ValueListWithoutDuplicates",valuesList.size()+"");
    int counter = 0;
    for(int i=0;i<valuesList.size();i++){
        for(int j=0;j<parent.size();j++){
            if(valuesList.get(i).equals(parent.get(j))){
                if(counter==0){
                    counter++;
                }else {
                    parent.set(j,parent.get(j)+String.valueOf(counter));
                    counter++;
                }
            }
        }


        counter=0;
    }

   for(int i=0;i<parent.size();i++){
       Log.d("final value",parent.get(i)+"");
   }