Sort with a Chain of Comparators

/*
 * Problem : Process lines.
 * Input   : Several lines.
 * Output  : The sorted lines.
 * Approach: Chained comparators.
 * Time    : O(n log n)
 * Space   : O(n)
 * Java    : 8 or newer.
 */
import java.util.Scanner;

public class Main {
        // Step 1: A Person keeps the three fields. Comparators decide the order.
        static class Person {
            final String name;
            final int age;
            final long score;

            Person(String name, int age, long score) {
                this.name = name;
                this.age = age;
                this.score = score;
            }
        }

    // This method solves the task. It receives the input and returns the answer.
    static String[] solve(String[] lines) {
        java.util.List<Person> people = new java.util.ArrayList<>();
        for (String line : lines) {
            String[] part = line.split(" ");
            people.add(new Person(part[0], Integer.parseInt(part[1]), Long.parseLong(part[2])));
        }
        // Step 2: Chain the comparators. The next one is used only when the earlier one says "equal".
        people.sort(java.util.Comparator
            .comparingLong((Person p) -> p.score).reversed()
            .thenComparingInt(p -> p.age)
            .thenComparing(p -> p.name));
        String[] output = new String[lines.length];
        for (int i = 0; i < output.length; i++) {
            Person p = people.get(i);
            output[i] = p.name + " " + p.age + " " + p.score;
        }
        return output;
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        java.util.List<String> all = new java.util.ArrayList<>();
        while (scanner.hasNextLine()) {
            String line = scanner.nextLine().trim();
            if (!line.isEmpty()) {
                all.add(line);
            }
        }
        String[] lines = all.toArray(new String[0]);
        System.out.println(String.join("\n", solve(lines)));
    }
}