队列的底层设计以及时间复杂度的优化

1.基于数组的实现

队列的基本概念就是先进先出，但是如果采用数组来实现就会出现每次执行一次dequeue会把数组里面的每个元素都会移动一次，移动n次的时间复杂度就是O（n^2），我们想要优化它的dequeue操作，可以定义两个变量来记录它的栈顶和栈尾来进行入队和出队操作

/**
 * Created by upupgogogo on 2018/4/26.下午6:02
 */
public class LoopQueue<E> implements Queue<E> {

    private E[] data;     //容器

    private int size;    //队列长度

    private int front;   //栈顶位置

    private int tail;   //栈尾位置

    public LoopQueue(int capacity){
        data = (E[]) new Object[capacity];
        size = 0;
        front = 0;
        tail = 0;
    }

    public LoopQueue(){
        this(10);
    }

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

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

    @Override
    public void enqueue(E e){
        if (size == getCapacity() ){
            resize(getCapacity() * 2);   // 扩容
        }
        data[tail] = e;
        tail = (tail + 1) % data.length;  //这个是一个循坏的队列
        size++;
    }

    @Override
    public E dequeue(){
        if(isEmpty())
            throw new IllegalArgumentException("Cannot dequeue from an empty queue.");

        E ret = data[front];
        data[front] = null;
        front = (front + 1) % data.length;
        size --;
        if(size == getCapacity() / 4 && getCapacity() / 2 != 0)
            resize(getCapacity() / 2);
        return ret;
    }

    @Override
    public E getFront(){
        if(isEmpty())
            throw new IllegalArgumentException("Cannot dequeue from an empty queue.");
        E ret = data[front];
        return ret;

    }

    public int getCapacity(){
        return data.length ;
    }


    private void resize(int capacity) {
        E[] newData = (E[]) new Object[capacity];
        for (int i = 0; i < size; i++)
            newData[i] = data[(i+front) % data.length];
        data = newData;
        front = 0;
        tail = size;

    }

    @Override
    public String toString(){

        StringBuilder res = new StringBuilder();
        res.append(String.format("Queue: size = %d , capacity = %d\n", size, getCapacity()));
        res.append("front [");
        for (int i = 0; i < size; i++){
            res.append(data[(i+front) % data.length]);
            if (i != size - 1)
                res.append(", ");
        }
        res.append("] tail");
        return res.toString();
    }

2.基于链表的实现

链表本身会有一个头指针，它进行dequeue时时间复杂度是O(1)，但是进行enqueue的操作却是O(n)级别的，我们可以设置一个tail变量来记录尾结点，这样enqueue操作也是O(1)级别的

/**
 * Created by upupgogogo on 2018/5/9.下午7:43
 */
public class LinkedListQueue<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() {
        this.head = null;
        this.size = 0;
        this.tail = null;
    }

    public int getSize(){
        return this.size;
    }

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

    public void enqueue(E e){
       if (isEmpty()){     //如果队列是空的，那么设置tail和head都指向new Node（e）
           tail = new Node(e);
           head = tail;
       }
       else {
           tail.next = new Node(e);
           tail = tail.next;
       }
       size ++;

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

    public E getFront(){
        if (isEmpty()){
            throw new IllegalArgumentException("Cannot getFront from an empty queue.");
        }
        return head.e;
    }

    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();
    }
}