Question

所以我最近了解了Generics并认为在优先级队列中实现它们会很酷。我有一个“Block”元素，它有一个firstName和一个数据变量。优先级队列的节点由Block，Next和Prev。

组成

我附上以下代码。我几乎完全得到“应该参数化”的错误/警告。并且一个错误表明我的“数据”元素无法解析为字段，这可能意味着我无法告诉我希望Block作为节点中的“元素E”。任何建议都将深表感谢

package QGen;

public class Block<E> implements Comparable<Block<E>> {
    protected String firstName;
    protected int data;

    public Block(String firstName, int data) {
        super();
        this.firstName = firstName;
        this.data = data;

    }

    @Override
    public int compareTo(Block x) {
        return (this.data - x.data);
    }
}


package QGen;

public class PriorityQueue<E extends Comparable> {
    protected Node<E> firstSentinel;
    protected Node<E> lastSentinel;

    protected class Node<E> {
        protected Node<E> next;
        protected Node<E> prev;
        private E element;

        public Node(E e, Node<E> previous, Node<E> nextt) {
            element = e;
            prev = previous;
            next = nextt;
        }
    }

    public PriorityQueue() {
        firstSentinel = new Node<>(null, null, null);
        lastSentinel = new Node<>(null, null, null);
        firstSentinel.data = 11111;
        lastSentinel.data = 0;
        firstSentinel.prev = null;
        firstSentinel.next = lastSentinel;
        lastSentinel.prev = firstSentinel;
        lastSentinel.next = null;
    }

    public void enQueue(E x) {
        Node<E> newX = new Node<E>(x, null, null);
        if (firstSentinel.next == lastSentinel)// list is empyty
        {
            firstSentinel.next = newX;
            newX.prev = firstSentinel;
            newX.next = lastSentinel;
            lastSentinel.prev = newX;
        } else {
            Node<E> temp = newX;
            Node<E> curr = firstSentinel.next;
            while (curr != lastSentinel && temp.element.compareTo(curr) <= 0) {// <=comparison
                // replaced
                curr = curr.next;
            }
            Node<E> tempCurr = curr;
            temp.next = tempCurr;
            temp.prev = tempCurr.prev;
            tempCurr.prev.next = temp;
            tempCurr.prev = temp;

        }
    }

    public E deQueue() {
        if (firstSentinel.next == lastSentinel) {
            return null;
        } else {
            Node<E> temp = new Node<E>(null, null, null);
            temp = firstSentinel.next;
            firstSentinel.next = temp.next;
            temp.next.prev = firstSentinel;
            return temp.element;
        }
    }

    public void printt() {
        Node<E> temp = new Node<E>(null, null, null);
        temp = firstSentinel.next;
        while (temp != lastSentinel) {
            System.out
                    .println(temp.element.firstName + " " + temp.element.data);

            temp = temp.next;
        }
    }
}

package QGen;

public class containsMain<E> {

    public static void main(String[] args) {

        PriorityQueue<Block> example = new PriorityQueue<Block>();
        Block dequedObject = new Block<>(null, null);
        Block<Block> incomingName = new Block<>("r", 1);
        example.enQueue(incomingName);
        dequedObject = (Block) example.deQueue();

    }

}

我知道我的PriorityQueue可能不是最好的实现，我会改进它。这是泛型，我无法提出解决方案

由于

Answer 1

在不查看方法逻辑的情况下，我有几个关于泛型的评论：

为什么Block本身是通用的？它不包含任何通用字段，因此将其从Block中删除！

new Node<>(null, null, null);
// this is a bad idea, change it to new Node<E>(null, null, null)


firstSentinel.data = 11111;
lastSentinel.data = 0;

//Those two cant work, because firstSentinel is referencing a Node<E> and not a Block! Delete those two rows, as they make no sense in your generic implementation of the Queue

Block<Block> incomingName = new Block<>("r", 1); 
// This doesnt make sense, should be Block incomingName = new Block(...)

优先级队列通用

1 个答案: