Object Ordering

A List l may be sorted as follows.

Collections.sort(l);

If the List consists of String elements, it will be sorted into alphabetical order. If it consists of Date elements, it will be sorted into chronological order. How does this happen?String and Date both implement the Comparable interface.Comparable implementations provide a natural ordering for a class, which allows objects of that class to be sorted automatically.The following table summarizes some of the more important Java platform classes that implement Comparable.

**Classes Implementing Comparable**
Class	Natural Ordering
`Byte`	Signed numerical
`Character`	Unsigned numerical
`Long`	Signed numerical
`Integer`	Signed numerical
`Short`	Signed numerical
`Double`	Signed numerical
`Float`	Signed numerical
`BigInteger`	Signed numerical
`BigDecimal`	Signed numerical
`Boolean`	`Boolean.FALSE < Boolean.TRUE`
`File`	System-dependent lexicographic on path name
`String`	Lexicographic
`Date`	Chronological
`CollationKey`	Locale-specific lexicographic

If you try to sort a list, the elements of which do not implement Comparable,Collections.sort(list) will throw aClassCastException. Similarly, Collections.sort(list, comparator) will throw a ClassCastException if you try to sort a list whose elements cannot be compared to one another using thecomparator. Elements that can be compared to one another are called mutually comparable. Although elements of different types may be mutually comparable, none of the classes listed here permit interclass comparison.

This is all you really need to know about the Comparable interface if you just want to sort lists of comparable elements or to create sorted collections of them. The next section will be of interest to you if you want to implement your own Comparable type.

Writing Your Own Comparable Types

The Comparable interface consists of the following method.

public interface Comparable<T> {
    public int compareTo(T o);
}

The compareTo method compares the receiving object with the specified object and returns a negative integer, 0, or a positive integer depending on whether the receiving object is less than, equal to, or greater than the specified object. If the specified object cannot be compared to the receiving object, the method throws aClassCastException.

The following class representing a person's name implements Comparable.

import java.util.*;

public class Name implements Comparable<Name> {
    private final String firstName, lastName;

    public Name(String firstName, String lastName) {
        if (firstName == null || lastName == null)
            throw new NullPointerException();
        this.firstName = firstName;
        this.lastName = lastName;
    }

    public String firstName() { return firstName; }
    public String lastName()  { return lastName;  }

    public boolean equals(Object o) {
        if (!(o instanceof Name))
            return false;
        Name n = (Name) o;
        return n.firstName.equals(firstName) && n.lastName.equals(lastName);
    }

    public int hashCode() {
        return 31*firstName.hashCode() + lastName.hashCode();
    }

    public String toString() {
	return firstName + " " + lastName;
    }

    public int compareTo(Name n) {
        int lastCmp = lastName.compareTo(n.lastName);
        return (lastCmp != 0 ? lastCmp : firstName.compareTo(n.firstName));
    }
}

To keep the preceding example short, the class is somewhat limited: It doesn't support middle names, it demands both a first and a last name, and it is not internationalized in any way. Nonetheless, it illustrates the following important points:

Name objects are immutable. All other things being equal,immutable types are the way to go, especially for objects that will be used as elements inSets or as keys in Maps. These collections will break if you modify their elements or keys while they're in the collection.
The constructor checks its arguments for null. This ensures that all Name objects are well formed so that none of the other methods will ever throw aNullPointerException.
The hashCode method is redefined. This is essential for any class that redefines theequals method. (Equal objects must have equal hash codes.)
The equals method returns false if the specified object isnull or of an inappropriate type. The compareTo method throws a runtime exception under these circumstances. Both of these behaviors are required by the general contracts of the respective methods.
The toString method has been redefined so it prints the Name in human-readable form. This is always a good idea, especially for objects that are going to get put into collections. The various collection types' toString methods depend on the toString methods of their elements, keys, and values.

Since this section is about element ordering, let's talk a bit more about Name's compareTo method. It implements the standard name-ordering algorithm, where last names take precedence over first names. This is exactly what you want in a natural ordering. It would be very confusing indeed if the natural ordering were unnatural!

Take a look at how compareTo is implemented, because it's quite typical.First, you compare the most significant part of the object(in this case, the last name). Often, you can just use the natural ordering of the part's type. In this case, the part is a String and the natural (lexicographic) ordering is exactly what's called for. If the comparison results in anything other than zero, which represents equality, you're done: You just return the result. If the most significant parts are equal, you go on to compare the next most-significant parts. In this case, there are only two parts — first name and last name. If there were more parts, you'd proceed in the obvious fashion, comparing parts until you found two that weren't equal or you were comparing the least-significant parts, at which point you'd return the result of the comparison.

Just to show that it all works, here's a program that builds a list of names and sorts them.

import java.util.*;

public class NameSort {
    public static void main(String[] args) {
        Name nameArray[] = {
            new Name("John", "Smith"),
            new Name("Karl", "Ng"),
            new Name("Jeff", "Smith"),
            new Name("Tom", "Rich")
        };

        List<Name> names = Arrays.asList(nameArray);
        Collections.sort(names);
        System.out.println(names);
    }
}

If you run this program, here's what it prints.

[Karl Ng, Tom Rich, Jeff Smith, John Smith]

There are four restrictions on the behavior of the compareTo method, which we won't go into now because they're fairly technical and boring and are better left in the API documentation. It's really important that all classes that implementComparable obey these restrictions, so read the documentation for Comparable if you're writing a class that implements it. Attempting to sort a list of objects that violate the restrictions has undefined behavior. Technically speaking, these restrictions ensure that the natural ordering is atotal order on the objects of a class that implements it; this is necessary to ensure that sorting is well defined.

Comparators

What if you want to sort some objects in an order other than their natural ordering? Or what if you want to sort some objects that don't implementComparable? To do either of these things, you'll need to provide a Comparator — an object that encapsulates an ordering. Like the Comparable interface, the Comparator interface consists of a single method.

public interface Comparator<T> {
    int compare(T o1, T o2);
}

The compare method compares its two arguments, returning a negative integer, 0, or a positive integer depending on whether the first argument is less than, equal to, or greater than the second. If either of the arguments has an inappropriate type for the Comparator, the compare method throws a ClassCastException.

Much of what was said about Comparable applies to Comparator as well.Writing acompare method is nearly identical to writing acompareTo method, except that the former gets both objects passed in as arguments. Thecompare method has to obey the same four technical restrictions as Comparable's compareTo method for the same reason — a Comparator must induce a total order on the objects it compares.

Suppose you have a class called Employee, as follows.

public class Employee implements Comparable<Employee> {
    public Name name()     { ... }
    public int number()    { ... }
    public Date hireDate() { ... }
       ...
}

Let's assume that the natural ordering of Employee instances is Name ordering (as defined in the previous example) on employee name. Unfortunately, the boss has asked for a list of employees in order of seniority. This means we have to do some work, but not much. The following program will produce the required list.

import java.util.*;
public class EmpSort {
    static final Comparator<Employee> SENIORITY_ORDER = 
                                        new Comparator<Employee>() {
            public int compare(Employee e1, Employee e2) {
                return e2.hireDate().compareTo(e1.hireDate());
            }
    };

    // Employee database
    static final Collection<Employee> employees = ... ;

    public static void main(String[] args) {
        List<Employee> e = new ArrayList<Employee>(employees);
        Collections.sort(e, SENIORITY_ORDER);
        System.out.println(e);
    }
}

The Comparator in the program is reasonably straightforward. It relies on the natural ordering of Date applied to the values returned by the hireDate accessor method. Note that the Comparator passes the hire date of its second argument to its first rather than vice versa. The reason is that the employee who was hired most recently is the least senior; sorting in the order of hire date would put the list in reverse seniority order. Another technique people sometimes use to achieve this effect is to maintain the argument order but to negate the result of the comparison.

// Don't do this!!
return -r1.hireDate().compareTo(r2.hireDate());

You should always use the former technique in favor of the latter because the latter is not guaranteed to work. The reason for this is that the compareTo method can return any negative int if its argument is less than the object on which it is invoked. There is one negative int that remains negative when negated, strange as it may seem.

-Integer.MIN_VALUE == Integer.MIN_VALUE

The Comparator in the preceding program works fine for sorting aList, but it does have one deficiency: It cannot be used to order a sorted collection, such asTreeSet, because it generates an ordering that isnot compatible with equals. This means that this Comparator equates objects that theequals method does not. In particular, any two employees who were hired on the same date will compare as equal. When you're sorting aList, this doesn't matter; but when you're using the Comparator to order a sorted collection, it's fatal. If you use thisComparator to insert multiple employees hired on the same date into aTreeSet, only the first one will be added to the set; the second will be seen as a duplicate element and will be ignored.

To fix this problem, simply tweak the Comparator so that it produces an ordering thatis compatible with equals. In other words, tweak it so that the only elements seen as equal when using compare are those that are also seen as equal when compared usingequals. The way to do this is to perform a two-part comparison (as forName), where the first part is the one we're interested in — in this case, the hire date — and the second part is an attribute that uniquely identifies the object. Here the employee number is the obvious attribute. This is theComparator that results.

static final Comparator<Employee> SENIORITY_ORDER = 
                                        new Comparator<Employee>() {
    public int compare(Employee e1, Employee e2) {
        int dateCmp = e2.hireDate().compareTo(e1.hireDate());
        if (dateCmp != 0)
            return dateCmp;

        return (e1.number() < e2.number() ? -1 :
               (e1.number() == e2.number() ? 0 : 1));
    }
};

One last note: You might be tempted to replace the final return statement in the Comparator with the simpler:

return e1.number() - e2.number();

Don't do it unless you're absolutely sure no one will ever have a negative employee number! This trick does not work in general because the signed integer type is not big enough to represent the difference of two arbitrary signed integers. Ifi is a large positive integer and j is a large negative integer,i - j will overflow and will return a negative integer. The resulting comparator violates one of the four technical restrictions we keep talking about (transitivity) and produces horrible, subtle bugs. This is not a purely theoretical concern; people get burned by it.

The SortedSet Interface

A SortedSet is a Set that maintains its elements in ascending order, sorted according to the elements' natural ordering or according to aComparator provided at SortedSet creation time. In addition to the normal Set operations, the SortedSet interface provides operations for the following:

Range view — allows arbitrary range operations on the sorted set
Endpoints — returns the first or last element in the sorted set
Comparator access — returns the Comparator, if any, used to sort the set

The code for the SortedSet interface follows.

public interface SortedSet<E> extends Set<E> {
    // Range-view
    SortedSet<E> subSet(E fromElement, E toElement);
    SortedSet<E> headSet(E toElement);
    SortedSet<E> tailSet(E fromElement);

    // Endpoints
    E first();
    E last();

    // Comparator access
    Comparator<? super E> comparator();
}

Set Operations

The operations that SortedSet inherits from Set behave identically on sorted sets and normal sets with two exceptions:

The Iterator returned by the iterator operation traverses the sorted set in order.
The array returned by toArray contains the sorted set's elements in order.

Although the interface doesn't guarantee it, the toString method of the Java platform'sSortedSet implementations returns a string containing all the elements of the sorted set, in order.

Standard Constructors

By convention, all general-purpose Collection implementations provide a standard conversion constructor that takes aCollection; SortedSet implementations are no exception. InTreeSet, this constructor creates an instance that sorts its elements according to their natural ordering. This was probably a mistake. It would have been better to check dynamically to see whether the specified collection was aSortedSet instance and, if so, to sort the new TreeSet according to the same criterion (comparator or natural ordering).Because TreeSet took the approach that it did, it also provides a constructor that takes aSortedSet and returns a new TreeSet containing the same elements sorted according to the same criterion. Note that it is the compile-time type of the argument, not its runtime type, that determines which of these two constructors is invoked (and whether the sorting criterion is preserved).

SortedSet implementations also provide, by convention, a constructor that takes aComparator and returns an empty set sorted according to the specifiedComparator. If null is passed to this constructor, it returns a set that sorts its elements according to their natural ordering.

Range-view Operations

The range-view operations are somewhat analogous to those provided by theList interface, but there is one big difference. Range views of a sorted set remain valid even if the backing sorted set is modified directly. This is feasible because the endpoints of a range view of a sorted set are absolute points in the element space rather than specific elements in the backing collection, as is the case for lists. A range-view of a sorted set is really just a window onto whatever portion of the set lies in the designated part of the element space. Changes to therange-view write back to the backing sorted set and vice versa. Thus, it's okay to userange-views on sorted sets for long periods of time, unlike range-views on lists.

Sorted sets provide three range-view operations. The first, subSet, takes two endpoints, like subList. Rather than indices, the endpoints are objects and must be comparable to the elements in the sorted set, using theSet's Comparator or the natural ordering of its elements, whichever theSet uses to order itself. Like subList, the range is half open, including its low endpoint but excluding the high one.

Thus, the following line of code tells you how many words between "doorbell" and"pickle", including "doorbell" but excluding "pickle", are contained in aSortedSet of strings called dictionary:

int count = dictionary.subSet("doorbell", "pickle").size();

In like manner, the following one-liner removes all the elements beginning with the letter f.

dictionary.subSet("f", "g").clear();

A similar trick can be used to print a table telling you how many words begin with each letter.

for (char ch = 'a'; ch <= 'z'; ) {
    String from = String.valueOf(ch++);
    String to = String.valueOf(ch);
    System.out.println(from + ": " + dictionary.subSet(from, to).size());
}

Suppose you want to view a closed interval, which contains both of its endpoints, instead of an open interval. If the element type allows for the calculation of the successor of a given value in the element space, merely request thesubSet from lowEndpoint to successor(highEndpoint). Although it isn't entirely obvious, the successor of a strings in String's natural ordering is s + "\0" — that is,s with a null character appended.

Thus, the following one-liner tells you how many words between "doorbell" and"pickle", including doorbell and pickle, are contained in the dictionary.

count = dictionary.subSet("doorbell", "pickle\0").size();

A similar technique can be used to view an open interval, which contains neither endpoint. The open-interval view from lowEndpoint to highEndpoint is the half-open interval from successor(lowEndpoint) to highEndpoint. Use the following to calculate the number of words between "doorbell" and "pickle", excluding both.

count = dictionary.subSet("doorbell\0", "pickle").size();

The SortedSet interface contains two morerange-view operations — headSet and tailSet, both of which take a singleObject argument. The former returns a view of the initial portion of the backing SortedSet, up to but not including the specified object. The latter returns a view of the final portion of the backing SortedSet, beginning with the specified object and continuing to the end of the backing SortedSet. Thus, the following code allows you to view the dictionary as two disjoint volumes ( a-m and n-z).

SortedSet<String> volume1 = dictionary.headSet("n");
SortedSet<String> volume2 = dictionary.tailSet("n");

Endpoint Operations

The SortedSet interface contains operations to return the first and last elements in the sorted set, not surprisingly calledfirst and last. In addition to their obvious uses, last allows a workaround for a deficiency in the SortedSet interface. One thing you'd like to do with aSortedSet is to go into the interior of the Set and iterate forward or backward. It's easy enough to go forward from the interior: Just get atailSet and iterate over it. Unfortunately, there's no easy way to go backward.

The following idiom obtains the first element that is less than a specified objecto in the element space.

Object predecessor = ss.headSet(o).last();

This is a fine way to go one element backward from a point in the interior of a sorted set. It could be applied repeatedly to iterate backward, but this is very inefficient, requiring a lookup for each element returned.

Comparator Accessor

The SortedSet interface contains an accessor method called comparator that returns the Comparator used to sort the set, ornull if the set is sorted according to the natural ordering of its elements. This method is provided so that sorted sets can be copied into new sorted sets with the same ordering. It is used by theSortedSet constructor described previously.

The SortedMap Interface

A SortedMap is a Map that maintains its entries in ascending order, sorted according to the keys' natural ordering, or according to aComparator provided at the time of the SortedMap creation. Natural ordering andComparators are discussed in the Object Ordering section. The SortedMap interface provides operations for normalMap operations and for the following:

Range view — performs arbitrary range operations on the sorted map
Endpoints — returns the first or the last key in the sorted map
Comparator access — returns the Comparator, if any, used to sort the map

The following interface is the Map analog of SortedSet.

public interface SortedMap<K, V> extends Map<K, V>{
    Comparator<? super K> comparator();
    SortedMap<K, V> subMap(K fromKey, K toKey);
    SortedMap<K, V> headMap(K toKey);
    SortedMap<K, V> tailMap(K fromKey);
    K firstKey();
    K lastKey();
}

Map Operations

The operations SortedMap inherits from Map behave identically on sorted maps and normal maps with two exceptions:

The Iterator returned by the iterator operation on any of the sorted map'sCollection views traverse the collections in order.
The arrays returned by the Collection views' toArray operations contain the keys, values, or entries in order.

Although it isn't guaranteed by the interface, the toString method of theCollection views in all the Java platform's SortedMap implementations returns a string containing all the elements of the view, in order.

Standard Constructors

By convention, all general-purpose Map implementations provide a standard conversion constructor that takes aMap; SortedMap implementations are no exception. In TreeMap, this constructor creates an instance that orders its entries according to their keys' natural ordering. This was probably a mistake. It would have been better to check dynamically to see whether the specifiedMap instance was a SortedMap and, if so, to sort the new map according to the same criterion (comparator or natural ordering).Because TreeMap took the approach it did, it also provides a constructor that takes aSortedMap and returns a new TreeMap containing the same mappings as the givenSortedMap, sorted according to the same criterion. Note that it is the compile-time type of the argument, not its runtime type, that determines whether theSortedMap constructor is invoked in preference to the ordinary map constructor.

SortedMap implementations also provide, by convention, a constructor that takes aComparator and returns an empty map sorted according to the specifiedComparator. If null is passed to this constructor, it returns aMap that sorts its mappings according to their keys' natural ordering.

Comparison to SortedSet

Because this interface is a precise Map analog of SortedSet, all the idioms and code examples inThe SortedSet Interface section apply to SortedMap with only trivial modifications.

Summary of Interfaces

The core collection interfaces are the foundation of the Java Collections Framework.

The Java Collections Framework hierarchy consists of two distinct interface trees:

The first tree starts with the Collection interface, which provides for the basic functionality used by all collections, such asadd and remove methods. Its subinterfaces — Set,List, and Queue — provide for more specialized collections.
The Set interface does not allow duplicate elements. This can be useful for storing collections such as a deck of cards or student records. TheSet interface has a subinterface, SortedSet, that provides for ordering of elements in the set.
The List interface provides for an ordered collection, for situations in which you need precise control over where each element is inserted. You can retrieve elements from aList by their exact position.
The Queue interface enables additional insertion, extraction, and inspection operations. Elements in aQueue are typically ordered in on a FIFO basis.
The Deque interface enables insertion, deletion, and inspection operations at both the ends. Elements in aDeque can be used in both LIFO and FIFO.
The second tree starts with the Map interface, which maps keys and values similar to aHashtable.
Map's subinterface, SortedMap, maintains its key-value pairs in ascending order or in an order specified by aComparator.

These interfaces allow collections to be manipulated independently of the details of their representation.

Answers to Questions and Exercises:

Questions

Question: At the beginning of this lesson, you learned that the core collection interfaces are organized into two distinctinheritance trees. One interface in particular is not considered to bea trueCollection, and therefore sits at the top of its own tree. What is the nameof this interface?
Answer: Map
Question: Each interface in the collections framework is declaredwith the<E> syntax, which tells you that it isgeneric. When you declare a Collection instance, what isthe advantage of specifying the type of objects that it will contain?
Answer: Specifying the type allows the compiler to verify (at compile time) that the type of object you put into the collection is correct, thus reducing errors at runtime.
Question: What interface represents a collection that does not allow duplicate elements?
Answer: Set
Question: What interface forms the root of the collections hierarchy?
Answer: Collection
Question: What interface represents an ordered collection that may contain duplicate elements?
Answer: List
Question: What interface represents a collection that holds elements prior to processing?
Answer: Queue
Question: What interface represents a type that maps keys to values?
Answer: Map
Question: What interface represents a double-ended queue?
Answer: Deque
Question: Name three different ways to iterate over the elements of aList.
Answer: You can iterate over aList using streams, the enhanced for statement, or iterators.
Question: True or False: Aggregate operations are mutative operations that modify the underlying collection.
Answer: False. Aggregate operations do not mutate the underlying collection. In fact, you must be careful to never mutate a collection while invoking its aggregate operations.Doing so could lead to concurrency problems should the stream be changed to a parallel stream at some point in the future.

Exercises

Exercise: Write a program that prints its arguments in random order. Do not make a copy of the argument array. Demonstrate how to print out the elements using both streams and the traditional enhanced for statement.
Answer:

import java.util.*;

public class Ran {

    public static void main(String[] args) {
        
        // Get and shuffle the list of arguments
        List<String> argList = Arrays.asList(args);
        Collections.shuffle(argList);

        // Print out the elements using JDK 8 Streams
        argList.stream()
        .forEach(e->System.out.format("%s ",e));

        // Print out the elements using for-each
        for (String arg: argList) {
            System.out.format("%s ", arg);
        }

        System.out.println();
    }
}

Exercise: Take the FindDups example and modify it to use a SortedSet instead of aSet. Specify a Comparator so that case is ignored when sorting and identifying set elements.
Answer:

import java.util.*;

public class FindDups {
    public static void main(String[] args) {
        Set<String> s = new HashSet<String>();
        for (String a : args)
               s.add(a);
               System.out.println(s.size() + " distinct words: " + s);
    }
}

Exercise: Write a method that takes a List<String> and appliesString.trim to each element.
Answer:
The enhanced for statement does not allow you to modify the List. Using an instance of theIterator class allows you to delete elements, but not replace an existing element or add a new one. That leavesListIterator:

import java.util.*;

public class ListTrim {
    static void listTrim(List<String> strings) {
        for (ListIterator<String> lit = strings.listIterator(); lit.hasNext(); ) {
            lit.set(lit.next().trim());
        }
    }

    public static void main(String[] args) {
        List<String> l = Arrays.asList(" red ", " white ", " blue ");
        listTrim(l);
        for (String s : l) {
            System.out.format("\"%s\"%n", s);
        }
    }
}

Exercise: Consider the four core interfaces, Set,List, Queue, and Map.For each of the following four assignments, specify which of the four core interfaces is best-suited, and explain how to use it to implement the assignment.
Answers:
- Whimsical Toys Inc (WTI) needs to record the names of all its employees. Every month, an employee will be chosen at random from these records to receive a free toy.
  Use a List. Choose a random employee by picking a number between0 and size()-1.
- WTI has decided that each new product will be named after an employee — but only first names will be used, and each name will be used only once. Prepare a list of unique first names.
  Use a Set. Collections that implement this interface don't allow the same element to be entered more than once.
- WTI decides that it only wants to use the most popular names for its toys. Count up the number of employees who have each first name.
  Use a Map, where the keys are first names, and each value is a count of the number of employees with that first name.
- WTI acquires season tickets for the local lacrosse team, to be shared by employees. Create a waiting list for this popular sport.
  Use a Queue. Invoke add() to add employees to the waiting list, andremove() to remove them.

Original: http://docs.oracle.com/javase/tutorial/collections/interfaces/order.html

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Java集合框架官方教程(3)：SortedSet/SortedMap接口

Object Ordering

Writing Your Own Comparable Types

Comparators

The SortedSet Interface

Set Operations

Standard Constructors

Range-view Operations

Endpoint Operations

Comparator Accessor

The SortedMap Interface

Map Operations

Standard Constructors

Comparison to SortedSet

Summary of Interfaces

Answers to Questions and Exercises:

Questions

Exercises

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