Algorithms and OOD (CSC 207 2013F) : Readings
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Summary: We consider the essential features of priority queues, our third form of linear structure.
Prerequisites: Linear Structures, Arrays, Polymorphism, Inheritance, Stacks, Queues.
You've seen two kinds of linear structures. Stacks are linear structures with a “last in, first out” policy. Queues are linear structures with a “first in, first out” policy. But sometimes neither policy suffices. For example, if we're using a linear structure to keep track of tasks, we typically want neither the first task we added nor the most recent task. Rather, we want the highest priority task. Linear structures that implement the “highest priority first” policy are called priority queues.
How do we represent priority? Typically, with a Comparator. You can decide whether the “smallest” or “largest” value has the highest priority. (For our sample code, we've chosen to choose the smallest value, since that mimics the decision in the standard Java Priority Queue implementation.) Where does the comparator come from? We'll supply it as a parameter to the constructor. (Of course, that's a bit dodgy to describe in the interface, so we'll choose the hack of having the interface also mention the comparator.)
So, what methods do we expect a priority queue to provide? Mostly, just the linear structure methods, but with a new interpretation. We'll also include a method that grabs the Comparator (again, mostly so that we can reference it).
What's left? Well, we need to think about the order in which the iterator will iterate elements. We could require that they be iterated in priority order, but that presupposes an implementation. So, we'll just require that the iterator provide all the elements in some order.
Putting it all together, we get the following:
package taojava.util; import java.util.Comparator; import java.util.Iterator; /** * A linear structure that follows the "best first" policy. * * @author Samuel A. Rebelsky */ public interface PriorityQueue<T> extends LinearStructure<T> { /** * Add an element to the queue. * * @param val * the value to add. * @pre * !this.isFull() * @post * The queue now contains an additional copy of val. * @exception Exception * If the structure is full. */ public void put(T val) throws Exception; /** * Remove the highest-priority element that is still in the * queue. * * @return * val, a value. * @pre * !this.isEmpty() * @post * The structure contains one fewer copy of val. * @post * For all values, v, in the queue, * comparator().compare(val, v) <= 0 * @exception Exception * If the structure is empty. */ public T get() throws Exception; /** * Determine what element will next be removed by get. * * @return * val, a value. * @pre * !this.isEmpty() * @post * For all values, v, in the queue, * comparator().compare(val, v) <= 0 * @exception Exception * If the structure is empty. */ public T peek() throws Exception; /** * Determine if the structure is empty. */ public boolean isEmpty(); /** * Determine if the structure is full. */ public boolean isFull(); /** * Get an iterator that returns all of the elements in some order * (that is, not necessarily in priority order). */ public Iterator<T> iterator(); /** * Determine what comparator is used by this priority queue. */ public Comparator<T> comparator(); } // interface PriorityQueue<T>
Take a minute to think about how you might implement priority queues using arrays.
Are you done? Why don't you take another moment and reflect a bit more.
Now are you done? See if you can come up with a second approach.
Sorry for the delay, but we really wanted you to think about implementation before we gave you some approaches. In fact, there are two standard approaches to implementing priority queues with arrays:
In the first case, the put
method will need to spend
some effort making sure that the underlying array remains in order.
However, the get
method is trivial: just remove and
return the element at the end of the array.
In the second case, the put
method is easy: just shove
the element at the end of the array. However, the get
and peek
methods will need to search the array for the
smallest element. For get
, you will then want to swap
the smallest element to the end of the array and return and remove
it.
Which method is better? In the first case, we spend O(n) work putting the new element at the next place and O(1) work getting that element. In the second case, we spend O(n) work finding the smallest element and O(1) work adding elements. So, if we find that our program is dominated by addition (e.g., because we add more tasks than we can ever hope to accomplish), we'll find the second is better. However, if we expect to get every element and also do a lot of peeking, the first approach is better.
The linked implementation is not much different. We can store the elements in order (in this case, probably highest-priority at the front) or we can store them in an arbitrary order and search and delete when we need the smallest. Storing the elements in order is probably the easier of the two strategies.
If you have a priority queue, sorting is pretty easy: Add all of the elements, then read them back out of the queue in order.
What's the running time? Well, either put or get is O(n). And we call both operation O(n) times, so it's O(n^{2}). Hmmm ... can we implement priority queues in a new way that allows us to put and get faster? That's a topic for the future.
Primary: [Front Door] [Schedule] - [Academic Honesty] [Disabilities] [Email] [FAQ] [IRC] [Teaching & Learning]
Current: [Assignment] [EBoard] [Lab] [Outline] [Partners] [Reading]
Sections: [Assignments] [EBoards] [Examples] [Handouts] [Labs] [Outlines] [Partners] [Readings]
Reference: [Java 7 API] [Java Code Conventions]
Related Courses: [CSC 152 2006S (Rebelsky)] [CSC 207 2013S (Walker)] [CSC 207 2011S (Weinman)]
Misc: [SamR] [Glimmer Labs] [CS@Grinnell] [Grinnell] [Issue Tracker (Course)] [Issue Tracker (Textbook)]
Copyright (c) 2013 Samuel A. Rebelsky.
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