• 介面中的資料域只能是public static final，方法只能是public abstract
  由於這個原因，這些修飾也可以忽略。

資料域只能是static final的原因：
stackoverflow上：

An interface can’t have behavior or state because it is intended to specify only an interaction contract, no implementation details. No behavior is enforced by not allowing method/constructor bodies or static/instance initializing blocks. No state is enforced by only allowing static final fields.Therefore, the Class can have a state (static state), but the instance state is not inferred by the interface.

• CompareTo方法和equals方法

強烈建議CompareTo應該與equals保持一致。對於兩個物件o1和o2, 應該確保以下同時成立:
o1.equals(o2)為true
o1.compareTo(o2)==0

The natural ordering for a class C is said to be consistent with equals if and only if e1.compareTo(e2) == 0 has the same boolean value as e1.equals(e2) for every e1 and e2 of class C. Note that null is not an instance of any class, and e.compareTo(null) should throw a NullPointerException even though e.equals(null) returns false

stackoverflow上有人提了一個問題，與這個有點關係，
Java, Why it is implied that objects are equal if compareTo() returns 0?

問題：

import java.util.Comparator;
import java.util.HashSet;
import java.util.Objects;
import java.util.Set;
import java.util.TreeSet;

public class Person implements Comparable<Person
 
> {

private final String name;
private int height;

public Person(String name,
        int height) {
    this.name = name;
    this.height = height;
}

public int getHeight() {
    return height;
}

public void setHeight(int height) {
    this.height = height;
}

public String getName() {
    return name;
}

@Override
public int compareTo(Person o) {
    return Integer.compare(height, o.height);
}

public boolean equals(Object obj) {
    if (obj == null) {
        return false;
    }
    if (getClass() != obj.getClass()) {
        return false;
    }
    final Person other = (Person) obj;
    if (!Objects.equals(this.name, other.name)) {
        return false;
    }
    return true;
}

public int hashCode() {
    int hash = 5;
    hash = 13 * hash + Objects.hashCode(this.name);
    return hash;
}

public String toString() {
    return "Person{" + name + ", height = " + height + '}';
}

public static class PComparator1 implements Comparator<Person> {

    @Override
    public int compare(Person o1,
            Person o2) {
        return o1.compareTo(o2);
    }
}

public static class PComparator2 implements Comparator<Person> {

    @Override
    public int compare(Person o1,
            Person o2) {
        int r = Integer.compare(o1.height, o2.height);
        return r == 0 ? o1.name.compareTo(o2.name) : r;
    }
}

public static void test(Set<Person> ps) {
    ps.add(new Person("Ann", 150));
    ps.add(new Person("Jane", 150));
    ps.add(new Person("John", 180));
    System.out.println(ps.getClass().getName());
    for (Person p : ps) {
        System.out.println(" " + p);
    }
}

public static void main(String[] args) {
    test(new HashSet<Person>());
    test(new TreeSet<Person>());
    test(new TreeSet<>(new PComparator1()));
    test(new TreeSet<>(new PComparator2()));
}
}

result:

java.util.HashSet
 Person{Ann, height = 150}
 Person{John, height = 180}
 Person{Jane, height = 150}

java.util.TreeSet
 Person{Ann, height = 150}
 Person{John, height = 180}

java.util.TreeSet
 Person{Ann, height = 150}
 Person{John, height = 180}

java.util.TreeSet
 Person{Ann, height = 150}
 Person{Jane, height = 150}
 Person{John, height = 180}

別人的解答，總結一下有下面幾點
1、首先，HashSet會根據hash code排序，
而TreeSet則是與equals、hash code沒有關係，
它只會按照class本身的compare來決定
2、第一點即解釋了問題的輸出結果
3、由上面的例子更能提醒我們，最好是使
(x.compareTo(y)==0) == (x.equals(y))
否則可能會出現一些意想不到的錯誤。