手撕Java集合之简易版Deque(LinkedList)

news2024/10/5 17:16:45

在目前，许多互联网公司的面试已经要求能手撕集合源码，集合源码本身算是源码里比较简单的一部分，但是要在面试极短的10来分钟内快速写出一个简易版的源码还是比较麻烦的，很容易出现各种小问题。所以在平时就要注重这方面的联系。
以下是我自己写的一个简易双端队列，我没有实现List接口，因为里面要实现的函数方法太多了，所以只是挑了几个核心的代码来写，本质其实就是头插法和尾插法的结合。
代码主要有三个文件，分别是Node节点，Deque类和测试文件。
Node:

package org.example.collection;

import lombok.Data;

import java.util.HashMap;
import java.util.Map;

@Data
public class Node<T> {
    T var;
    Node<T> prev;
    Node<T> next;

    Node(Node<T> prev,T element, Node<T> next){
        this.var = element;
        this.next = next;
        this.prev = prev;
    }

}

Deque实现文件：

package org.example.collection;

public class DequeCode<E>{
    int size = 0;
    Node<E> first;
    Node<E> last;
    public DequeCode() {
        this.first = new Node<E>(null,null,null);
        this.last = new Node<E>(null,null,null);
        //first和last之间应该建立联系
        first.next = last;
        last.prev = first;
    }
    public int size() {
        return size;
    }

    public void addFirst(E element){
        //采用头插法来进行双端的插入
        Node<E> node = new Node<>(null,element,null);
        size++;
        if(first.next==null){
            first.next = node;
            node.prev = first;
            return ;
        }
        Node<E> temp = first.next;
        first.next = node;
        node.prev = first;
        node.next = temp;
        temp.prev = node;
    }
    public void addLast(E element){
        //和头插入相同的思路
        Node<E> node = new Node<>(null,element,null);
        Node<E> temp = last.prev;
        size++;
        if(last.prev==null){
            last.prev = node;
            node.next = last;
            return ;
        }
        last.prev = node;
        node.next = last;
        temp.next = node;
        node.prev = temp;
    }
    public void removeFirst() throws Exception {
        if(size == 0) throw new Exception("出现问题");
        Node<E> node = first.next;
        first.next = first.next.next;
        node.next.prev = first;
    }
    public void removeLast() throws Exception {
        if(size == 0) throw new Exception("出现问题");
        Node<E> node = last.prev;
        last.prev = last.prev.prev;
        node.prev.next = last;
    }
    public Node<E> peekFirst(){
        return first.next;
    }
    public Node<E> peekLast(){
        return last.prev;
    }
}

最后是测试文件：

package org.example.collection;

public class TestDeque {
    public static void main(String[] args) throws Exception {
        DequeCode<Integer> deque = new DequeCode<>();
        deque.addFirst(1);
        deque.addLast(2);
        System.out.println(deque.peekFirst().var);
        System.out.println(deque.peekLast().var);
        deque.addFirst(3);
        deque.addLast(4);
        System.out.println(deque.peekFirst().var);
        System.out.println(deque.peekLast().var);
        deque.removeFirst();
        deque.removeLast();
        System.out.println(deque.peekFirst().var);
        System.out.println(deque.peekLast().var);
    }
}