使用带尾指针的链表实现队列

使用带尾指针的链表实现队列

  1. 普通链表在头部删除和添加节点是快速的,但是在尾部操作时间复杂度过高。如果要实现队列,则带尾指针的链表可以解决在尾部操作时间复杂度的问题。入队与出队的时间复杂度都为O(1)。

  2. 具体实现代码

package LinkedList.LinkedListQueue;

/**
 * @ Description: 使用链表实现队列
 * @ Date: Created in 14:04 12/07/2018
 * @ Author: Anthony_Duan
 */
public class LinkedListQueue<E> implements Queue<E> {

    private class Node {
        public E e;
        public Node next;

        public Node(E e, Node next) {
            this.e = e;
            this.next = next;
        }

        public Node(E e) {
            this(e, null);
        }

        public Node() {
            this(null, null);
        }

        @Override
        public String toString() {
            return e.toString();
        }
    }

    private Node head, tail;
    private int size;

    public LinkedListQueue() {
        head = null;
        tail = null;
        size = 0;
    }

    @Override
    public int getSize() {
        return size;
    }

    @Override
    public boolean isEmpty() {
        return size == 0;
    }

    @Override
    public void enqueue(E e) {
        if (tail == null) {
            tail = new Node(e);
            head = tail;
        } else {
            tail.next = new Node(e);
            tail = tail.next;
        }
        size++;
    }


    @Override
    public E dequeue() {
        if (isEmpty()) {
            throw new IllegalArgumentException("Cannot dequeue from an empty queue.");
        }
        Node retNode = head;
        head = head.next;
        retNode.next = null;
        if (head == null) {
            tail = null;
        }
        size--;
        return retNode.e;
    }

    @Override
    public E getFront() {
        if (isEmpty()) {
            throw new IllegalArgumentException("Queue is empty.");
        }
        return head.e;
    }

    @Override
    public String toString() {
        StringBuilder res = new StringBuilder();
        res.append("Queue: front ");

        Node cur = head;
        while (cur != null) {
            res.append(cur + "->");
            cur = cur.next;
        }
        res.append("NULL tail");
        return res.toString();
    }


    public static void main(String[] args) {
        LinkedListQueue<Integer> queue = new LinkedListQueue<Integer>();
        for (int i = 0; i < 10; i++) {
            queue.enqueue(i);
            System.out.println(queue);

            if (i % 3 == 2) {
                queue.dequeue();
                System.out.println(queue);
            }
        }
    }
}

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转载自blog.csdn.net/xiaoduan_/article/details/81204303