优化滤波算法

Question

我目前正在开发一个带有带有aprox的RecyclerView的活动的Android应用程序。1k项(数据库查看器)。用户必须能够搜索这些项，因此我使用以下筛选器实现了搜索：

class PersonFilter extends Filter
{
    private PersonViewAdapter adapter;
    private Person[] originalPeople;

    public PersonFilter(PersonViewAdapter adapter, Person[] unfiltered)
    {
        super();
        this.adapter = adapter;
        this.originalPeople = unfiltered;
    }

    @Override
    protected FilterResults performFiltering(CharSequence constraint)
    {
        constraint = constraint.toString().toLowerCase();
        FilterResults results = new FilterResults();

        if (constraint.length() == 0)
        {
            results.values = originalPeople;
            results.count = originalPeople.length;
        }
        else
        {
            ArrayList<Person> filtered = new ArrayList<>();

            for (Person p : originalPeople)
            {
                //region filtering
                if (p instanceof Student)
                {
                    Student s = (Student)p;
                    if (s.fullName.toLowerCase().contains(constraint) ||
                            s.mngClass.name.contains(constraint) ||
                            s.getShortUsername().contains(constraint) ||
                            s.getLongUsername().contains(constraint))
                    {
                        filtered.add(p);
                    }
                }
                else if (p instanceof Teacher)
                {
                    Teacher t = (Teacher)p;
                    if (t.fullName.toLowerCase().contains(constraint) ||
                            t.acronym.toLowerCase().contains(constraint) ||
                            t.getShortUsername().contains(constraint) ||
                            t.getLongUsername().contains(constraint))
                    {
                        filtered.add(p);
                    }
                }
                else
                {
                    if (p.fullName.toLowerCase().contains(constraint) ||
                            p.getShortUsername().contains(constraint) ||
                            p.getLongUsername().contains(constraint))
                    {
                        filtered.add(p);
                    }
                }
                //endregion
            }

            Person[] dummy = new Person[filtered.size()];
            results.values = filtered.toArray(dummy);
            results.count = filtered.size();
        }

        return results;
    }

    @Override
    protected void publishResults(CharSequence constraint, FilterResults results)
    {
        adapter.setPeople((Person[])results.values);
        adapter.notifyDataSetChanged();
    }
}

问题是，这样的搜索效果不佳，而且由于我在Java/Android性能优化方面经验不足，我想知道我在这里可以改进什么。

这里是'Person‘类(如果需要的话)：

public abstract class Person
{
    public int id;
    public String fullName;
    public String firstName;
    public String lastName;

    public String getShortUsername()
    {
        String ln = toUsernameForm(lastName);
        String username = ln.substring(0, Math.min(ln.length(), 7)) + firstName.substring(0, 1).toLowerCase();
        return username;
    }

    public String getLongUsername()
    {
        return toUsernameForm(firstName) + "." + toUsernameForm(lastName);
    }

    public static String toUsernameForm(String s)
    {
        return normalize(s.split("\\W")[0].toLowerCase().replace(" ", "")
                                        .replace("ä", "ae")
                                        .replace("ö", "oe")
                                        .replace("ü", "ue")
                                        .replace("ß", "ss"));
    }

    public static String normalize(String s)
    {
        String nfdNormalizedString = Normalizer.normalize(s, Normalizer.Form.NFD);
        Pattern pattern = Pattern.compile("\\p{InCombiningDiacriticalMarks}+");
        return pattern.matcher(nfdNormalizedString).replaceAll("");
    }
}

Answer 1

此版本应减少类型铸造。

ArrayList<Person> filtered = new ArrayList<>();

for (Person p : originalPeople)
{
    //region filtering
    if (p.fullName.toLowerCase().contains(constraint) ||
        p.getShortUsername().contains(constraint) ||
        p.getLongUsername().contains(constraint))
    {
        filtered.add(p);
    }
    else if (p instanceof Student)
    {
        Student s = (Student) p;
        if (s.mngClass.name.contains(constraint))
        {
            filtered.add(p);
        }
    }
    else if (p instanceof Teacher)
    {
        Teacher t = (Teacher) p;
        if (t.acronym.toLowerCase().contains(constraint))
        {
            filtered.add(p);
        }
    }
    //endregion
}

results.values = filtered.toArray();
results.count = filtered.size();

看起来，如果long用户名包含一个字符序列，那么短版本也可以(可能是特殊字符除外)。您可以利用这一事实来减少比较。