java - 尝试以不同的方式实现优先级队列

Question

我正在尝试实现一个优先级队列，它的功能与普通优先级队列有点不同。这里的想法是支持快速插入，但对最高优先级项目的访问速度较慢。在这种情况下，Insert 方法将新项目放在方便的地方；因此，remove 和 peek 方法必须在整个队列中搜索最高优先级的项目。

public class MinPriorityQueue {

/** Constructs a new MinPriorityQueue with capacity cap and size 0.
 * 
 * @param cap Capacity of the new queue.
 */
public MinPriorityQueue(int cap) {         

    this.queue = (Comparable []) new Comparable[cap];
    this.size = 0;
    this.cap = cap;
}

int size() {
   return this.size;
}

int capacity() {
   return this.cap;
}

 boolean isEmpty() {
   return this.size==0;
}

boolean isFull() {
    return this.size == cap;       
}

void insert(Comparable e) {
    if (this.size == cap) {
        throw new IllegalStateException();
    }
    queue[size] = e;
    size++;



Comparable remove() {

    if (this.size == 0) {
        throw new IllegalStateException();
    }
    int maxIndex = 0;
    for (int i=1; i<size; i++) { 
        if (queue[i].compareTo (queue[maxIndex]) < 0) { 
            maxIndex = i; 
        } 
    } 
    Comparable result = queue[maxIndex]; 
    size--; 
    queue[maxIndex] = queue[size]; 
    return result;
    }


 /** Returns, but does not remove, the lowest-priority item in the
 * queue.
 * 
 * Complexity: O(n)
 * @return Lowest priority item.
 * @throws IllegalStateException if the queue is empty.
 */
Comparable top()  {
           if (this.size == 0) {
        throw new IllegalStateException();
    }
         /*  int i;
           for (i=0; i < size; i++) {
        if (queue[i].compareTo (size-1) < 0) {
            break;
        }
    }
           return queue[size - 1];*/
    return  queue[size-1]; 
  }



Comparable[] queue; // Contents of the queue

private  int cap;
private int size;    

}

////测试文件

 /**
     * Test of remove method, of class MinPriorityQueue.
     */
    @Test
    public void testRemove() {
    System.out.println("remove");
    MinPriorityQueue q = new MinPriorityQueue(10);

    boolean throws_exception = false;
    try {
        q.remove();
    } catch (IllegalStateException e) {
        throws_exception = true;
    } catch (Throwable e) {

    }
    assertTrue("remove throws an exception when empty", throws_exception);

    // Permutation of 0...9
    int[] input = {0, 5, 9, 2, 3, 1, 6, 8, 7, 4};

    for (int i : input) {
        q.insert(i);
    }

    assertTrue(q.isFull());

    for (int i = 10; i > 0; --i) {
        assertEquals(q.size(), i);
        Integer x = (Integer) q.remove();
        assertEquals(x, new Integer(10 - i)); // Items are removed in correct order            
    }

    assertTrue(q.isEmpty());
}

/**
 * Test of top method, of class MinPriorityQueue.
 */
@Test
public void testTop() {
    System.out.println("top");
    MinPriorityQueue q = new MinPriorityQueue(10);

    boolean throws_exception = false;
    try {
        q.top();
    } catch (IllegalStateException x) {
        throws_exception = true;
    } catch (Throwable x) {

    }
    assertTrue("top throws an exception when empty", throws_exception);

    int[] input = {0, 5, 9, 2, 3, 1, 6, 8, 7, 4};
    int smallest = 10;
    for (int i : input) {
        q.insert(i);
        if (i < smallest) {
            smallest = i;
        }

        assertEquals(q.top(), smallest);
    }

    for (int i = 0; i < 10; ++i) {
        assertEquals(q.top(), i);
        q.remove();
    }
}    

我已经能够实现除 remove 和 peek 之外的所有方法，因为我无法获得实现这些方法的正确逻辑。我无法弄清楚我们如何搜索整个队列以找到最高优先级的项目。对于这个，我认为应该使用 for 循环，但逻辑不正确

编辑:我能够获得 remove() 方法的正确逻辑，只需要让 pop() 方法工作

Answer 1

实现此目的的最简单方法是使用 java.util.List而不是 array .所以 List会将元素保留在队列中并处理它们的定位而不对它们进行排序(因此它仍然使用快速插入)。如果你使用实现 java.util.ArrayList对于 List它使用 array 的界面在内部，所以这几乎正是你正在尝试的，但更容易，因为其他人已经完成了大部分工作......

所以你可以实现类MinPriorityQueue像这样:

import java.util.ArrayList;
import java.util.List;

//use a generic type argument T here that extends comparable so you can only store objects of a specific type that are comparable to each other
public class MinPriorityQueue<T extends Comparable<T>> {

    private List<T> queue; // Contents of the queue

    private int cap;
    //private int size;

    /**
     * Constructs a new MinPriorityQueue with capacity cap and size 0.
     * 
     * @param cap
     *        Capacity of the new queue.
     */
    public MinPriorityQueue(int cap) {
        this.queue = new ArrayList<T>(cap);
        //this.size = 0;
        this.cap = cap;
    }

    public int size() {
        //return this.size;
        return queue.size();
    }

    public int capacity() {
        return this.cap;
    }

    public boolean isEmpty() {
        //return this.size == 0;
        return queue.isEmpty();
    }

    public boolean isFull() {
        //return this.size == cap;
        return queue.size() == cap;
    }

    public void insert(T e) {
        if (queue.size() == cap) {
            throw new IllegalStateException();
        }
        //queue[size] = e;
        //size++;
        queue.add(e);
    }

    /**
     * Removes and returns the lowest-priority item from the queue.
     * 
     * Complexity: O(n)
     * 
     * @return Lowest priority item that was in the queue.
     * @throws IllegalStateException
     *         if the queue is empty.
     */
    public Comparable<T> remove() {
        if (queue.size() == 0) {
            throw new IllegalStateException();
        }

        /*for (int i = 0; i < this.size; i++) {
            if (this.queue[i] == queue[size]) {
                return i;
            }
        }
        return queue[--size];*/

        //initialize the lowest priority item with the first one in the list
        int lowestPriorityIndex = 0;

        //search every other item in the list to see whether it has a lower priority than the current lowest priority 
        for (int i = 1; i < queue.size(); i++) {
            if (queue.get(i).compareTo(queue.get(lowestPriorityIndex)) < 0) {
                lowestPriorityIndex = i;
            }
        }

        //return and remove the lowest priority item
        return queue.remove(lowestPriorityIndex);
    }

    /**
     * Returns, but does not remove, the lowest-priority item in the queue.
     * 
     * Complexity: O(n)
     * 
     * @return Lowest priority item.
     * @throws IllegalStateException
     *         if the queue is empty.
     */
    public Comparable<T> top() {
        if (queue.size() == 0) {
            throw new IllegalStateException();
        }
        /*  int i;
          for (i=0; i < size; i++) {
        if (queue[i].compareTo (size-1) < 0) {
           break;
        }
        }
          return queue[size - 1];*/


        //initialize the lowest priority item with the first one in the list
        int lowestPriorityIndex = 0;

        //search every other item in the list to see whether it has a lower priority than the current lowest priority 
        for (int i = 1; i < queue.size(); i++) {
            if (queue.get(i).compareTo(queue.get(lowestPriorityIndex)) < 0) {
                lowestPriorityIndex = i;
            }
        }

        //return (but not remove) the lowest priority item
        return queue.get(lowestPriorityIndex);
    }
}

此代码通过了您在问题中提供的所有测试用例。

我还在示例代码中添加了一个泛型类型参数(因此类声明现在是 MinPriorityQueue<T extends Comparable<T>> )，因为它需要实际比较元素。您可以检查泛型类型 here .