Kotlin 根据另一个列表的顺序对列表进行排序

我需要对Person个对象的列表进行排序(List<Person>，其中每Person个对象都有一些属性，比如id(唯一)、name、age……等等).

The sorting order is based on another list. That list contains a set of Person id's (A List<String> which is already sorted).

What is the best way to order the List<Person> in the same order as the list of id's using Kotlin or Java.

示例:

List Person {
(“ID1”,”PERSON1”,22,..), (“ID-2”,”PERSON2”,20,..) ), (“ID-3”,”PERSON3”,19,..),…..
}

Ordered Id List :

List of ID {(“ID2”), (“ID1”),(”ID3”)….}

Sorted Person list should be:

List PERSON {
 (“ID-2”,”PERSON 2”,20,..) ), (“ID1”,”PERSON 2”,22,..),  (“ID-3”,”PERSON 2”,19,..),…..
}

如果Person列表包含id列表中未提及的任何id，则这些值应位于排序列表的末尾.

Edited: This is my current way in Java. I am hoping for a better way than this:

public static List<Person> getSortList(List <Person> unsortedList, List<String> orderList){

    if(unsortedList!=null && !unsortedList.isEmpty() && orderList!=null && !orderList.isEmpty()){
        List sortedList = new ArrayList<OpenHABWidget>();
        for(String id : orderList){
            Person found= getPersonIfFound(unsortedList, id); // search for the item on the list by ID
            if(found!=null)sortedList.add(found);       // if found add to sorted list
            unsortedList.remove(found);        // remove added item
        }
        sortedList.addAll(unsortedList);        // append the reaming items on the unsorted list to new sorted list
        return sortedList;
    }
    else{
        return unsortedList;
    }

}

public static Person getPersonIfFound(List <Person> list, String key){
    for(Person person : list){
        if(person.getId().equals(key)){
            return person;
        }
    }
    return null;
}