What is the best way to sort lists in Java?

Question

I have a list, for example:
I want to know how best to sort that list by name and age respectively.

public class Pessoa{
    private String nome;
    private int idade;

    // gets e sets omitidos
}

public class PessoaNeg{
    private List<Pessoa> listaPessoas = new ArrayList<Pessoa>();

    public void adicionarPessoas(){
       Pessoa pessoa = new Pessoa();
       pessoa.setNome("João");
       pessoa.setIdade(10); 

       Pessoa pessoa = new Pessoa();
       pessoa.setNome("Filomena");
       pessoa.setIdade(11); 

       listaPessoas.add(pessoa);
    }

}

Answer 1

Okay, the best best shape will depend on many factors, including the conjunction of the astral projection of the Sun in relation to Venus and where its shadow projects in the house of the Lunar Signs...

A way suitable to make this ordination is calling the Collections.sort.

To solution from @Isac I believe it is well decoupled, but there is a situation where, if the object is guaranteed to be comparable and absolutely manageable, there is no need to implement a Comparator anonymous/ephemeral.

In this case, the object needs to implement Comparable. So how are we dealing with the class Pessoa and if you wish to compare it to yourself, let us do this:

public class Pessoa implements Comparable<Pessoa> {
    private String nome;
    private int idade;

    @Override
    public int compareTo(Pessoa outra) {
        int nomeCmp = nome.compareTo(outra.nome);
        // deu empate quanto aos nomes, então vai para o desempate de idade
        if (nomeCmp == 0) {
            return idade - outra.idade;
        }
        return nomeCmp;
    }

    // gets e sets omitidos
}

All right, now you just need to call someone who knows how to sort out a list of comparable... and that’s the Collections.sort(List<T extends Comparable<? super T>).

Now, just call:

List<Pessoa> pessoas = new ArrayList<>();

// ... povoa a lista ...

Collections.sort(pessoas);

PS: I would use normally the solution of @Isac, I usually do not have in many cases this absolute ordination.

Answer 2

There are many ways to sort, either by hand or already using done functions. And there are several types of sorting algorithms. Each can be better or worse depending on the purpose and type of data you have.

I will show you one using Amblas and based on the internal algorithm of sort for ArrayList. To order only by nome can do:

public class Main {
    public static void main(String[] args) {
       List<Pessoa> listaPessoas = new ArrayList<>();

       listaPessoas.add(new Pessoa("João", 10));
       listaPessoas.add(new Pessoa("Filomena", 11));
       listaPessoas.add(new Pessoa("Martim", 25));
       listaPessoas.add(new Pessoa("Ana", 21));
       listaPessoas.add(new Pessoa("Rui", 9));  

       listaPessoas.sort((p1,p2)->p1.getNome().compareTo(p2.getNome())); //ordenação
       listaPessoas.forEach(x->System.out.println(x.obterInformacoes()));
    }
}

Exit:

Nome: Ana, Idade: 21
Nome: Filomena, Idade: 11
Nome: João, Idade: 10
Nome: Martim, Idade: 15
Nome: Rui, Idade: 9

In which here he sees that the ordination of the people is based on the comparison of their names, since it is called the compareTo of the name of p1, the first person, to p2.

See this example in Ideone

If you want to sort by name and for each name equal sort by age you already have to elaborate a little in the sorting method:

listaPessoas.sort((p1,p2)-> {
    int compNomes = p1.getNome().compareTo(p2.getNome());
    return compNomes == 0 ? p1.getIdade()-p2.getIdade():compNomes;
});

Note that the compareTo return 0 when both Strings are the same.

See also this example in Ideone

Documentation:

• sort
• compareTo