用双链表实现双端队列

双端队列：这个结构允许你从头部加，头部出，允许你从尾部加，或者尾部出。所以是4个操作。而且每一个操作都是O(1)

单链表支持头部的加减。

单链表也支持尾部的加，但是不支持尾部的减。因为tail无法找到它的之前的节点。可以通过遍历到前节点，但是不能遍历，因为要求时间复杂度是O(1)，所以单链表不支持双端队列。

 

双链表就可以：

 

代码实现：

package com.cy.class04;

/**
 * @description com.cy.class04
 * @author: chengyu
 * @date: 2025-05-01 11:08
 */
public class Code03_DoubleLinkedListToDeque {

    /**
     * 双向链表
     */
    public static class Node<V> {
        public V value;
        public Node<V> last;
        public Node<V> next;

        public Node(V v) {
            this.value = v;
            last = null;
            next = null;
        }
    }

    /**
     * 双端队列的实现
     */
    public static class MyDeque<V> {
        private Node<V> head;
        private Node<V> tail;
        private int size;

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

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

        public int size() {
            return size;
        }

        public void pushHead(V value) {
            Node<V> cur = new Node<>(value);
            if (head == null) {
                head = cur;
                tail = cur;
            } else {
                cur.next = head;
                head.last = cur;
                head = cur;
            }
            size++;
        }

        public void pushTail(V value) {
            Node<V> cur = new Node<>(value);
            if (head == null) {
                head = cur;
                tail = cur;
            } else {
                tail.next = cur;
                cur.last = tail;
                tail = cur;
            }
            size++;
        }

        public V poolHead() {
            V ans = null;
            if (head == null) {
                return ans;
            }
            size--;
            ans = head.value;
            if (head == tail) {
                head = null;
                tail = null;
            } else {
                head = head.next;
                head.last = null;
            }
            return ans;
        }

        public V poolTail() {
            V ans = null;
            if (head == null) {
                return ans;
            }
            size--;
            ans = tail.value;
            if (head == tail) {
                head = null;
                tail = null;
            } else {
                tail = tail.last;
                tail.next = null;
            }
            return ans;
        }

        public V peekHead() {
            V ans = null;
            if (head != null) {
                ans = head.value;
            }
            return ans;
        }

        public V peekTail() {
            V ans = null;
            if (tail != null) {
                ans = tail.value;
            }
            return ans;
        }
    }
}

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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