从另一个对象列表中提取对象列表

Question

I have a List<User> users , where class User has one property username . I also have another List<User> activeUsers . Let users = activeUsers + inactiveUsers . Now I want to extract inactiveUsers from users based on username property. I solved this problem using two for-loops. I think this is not efficient way. so if anyone know how it can be done efficiently please let me know.

eg I have activeUsers[1,3] and users[1,2,3,4] and want to build inactiveUsers[2,4].

Answer 1

In Java, you can use the Collection interface's removeAll method.

// Create a couple ArrayList objects and populate them
// with some delicious fruits.
Collection firstList = new ArrayList() {{
    add("user1");
    add("user2");
}};

Collection secondList = new ArrayList() {{
    add("user1");
    add("user1");
    add("user3");
    add("user4");
}};

// Show the "before" lists
System.out.println("First List: " + firstList);
System.out.println("Second List: " + secondList);

// Remove all elements in firstList from secondList
secondList.removeAll(firstList);

// Show the "after" list
System.out.println("Result: " + secondList);

The above code will produce the following output:

First List: [user1, user2]
Second List: [user1, user2, user3, user4]
Result: [user3, user4]

Answer 2

Well, based on the restriction of modifying the User class and nature of your question there is no way of doing it lesser than o(n^2) with two lists (ie two loops). Of course if you could have two lists of each type your problem is done.

But since you cannot than the logic is : (assuming your structures are lists only)

Iterate users list (o(n)) :
   - Search the active users list for the current user (o(n)) 

No matter how you look at it you get o(n^2)

If you could modify User to have activation property you can easily reduce the problem to o(n) by a single search

Answer 3

You could store users in a Set based on if they're active or not, I would prefer a property on User but I guess you don't want that in this assignment of yours. Using this code your problem is solved in O(n) time. Assuming you haven't messed up equals/hashCode in User.

package com.stackoverflow.q15999468;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashSet;
import java.util.List;
import java.util.Set;

public class Users
{
    public static void main(String[] args)
    {
        List<User> users = Arrays.asList(u1,u2,u3,u4,u5);
        List<User> activeUsers = Arrays.asList(u3,u5);

        Set<User> activeUsersSet = new HashSet<User>(activeUsers);

        List<User> inactiveUsers = new ArrayList<User>();

        for(User user : users)
        {
            if(!activeUsersSet.contains(user))
            {
                inactiveUsers.add(user);
            }
        }
        System.out.println("Inactive users: " + inactiveUsers);
    }
}

Answer 4

This should work

List<User> inactiveUsers = new ArrayList<User>(users);
inactiveUsers.removeAll(activeUsers) ;