LinkedList 的实现原理

本文为博客园作者所写: 一寸HUI,个人博客地址:https://www.cnblogs.com/zsql/

简单的一个类就直接说了。LinkedList 的底层结构是一个带头/尾指针的双向链表,可以快速的对头/尾节点 进行操作,它允许插 入所有元素,包括 null。 相比数组(这里可以对比ArrayList源码分析进行查看),链表的特点就是在指定位置插入和删除元素的效率较高,但是查找的 效率就不如数组那么高了。如果熟悉双向链表这个数据结构,其实就很简单了,无非就是实现一些数据的添加,删除,查询,遍历等功能,双向链表的结构图如下:

 

 

每一个数据(节点)都包含3个部分,一个是数据本身item,一个是指向下一个节点的next指针,还有就是指向上一个节点的prev指针,另外,双向链表还有一个 first 指针,指向头节点,和 last 指针,指向尾节点。,在LinkedList类中通过私有的静态内部类Node作为每一个数据的封装。具体实现如下:

private static class Node<E> {  //这个类就是用来封装双向链表中的每一个数据,也是上图中的每一个框
        E item;
        Node<E> next;
        Node<E> prev;

        Node(Node<E> prev,E element,Node<E> next) {
            this.item = element;
            this.next = next;
            this.prev = prev;
        }
    }

接下看看LinkList类的定义:

public class LinkedList<E>
    extends AbstractSequentialList<E> //继承的类
    implements List<E>,Deque<E>,Cloneable,java.io.Serializable //实现的各种接口
{}

 

 

 接下来看看LinkedList这个类的一些属性:就三个属性,一个用来记录双向链表的大小,一个是first节点用来指向链表的头,last用来指向链表的尾

   transient int size = 0;

    /**
     * Pointer to first node.
     * Invariant: (first == null && last == null) ||
     *            (first.prev == null && first.item != null)
     */
    transient Node<E> first;

    
     * Pointer to last node.
     * Invariant: (first == null && last == null) ||
     *            (last.next == null && last.item != null)
     transient Node<E> last;

在看看构造方法:


     * Constructs an empty list.
     public LinkedList() { //空参构造
    }


     * Constructs a list containing the elements of the specified
     * collection,in the order they are returned by the collection's
     * iterator.
     *
     * @param  c the collection whose elements are to be placed into this list
     * @throws NullPointerException if the specified collection is null
     public LinkedList(Collection<? extends E> c) { //通过已有的集合进行构造
        this();
        addAll(c); //使用addAll()方法把集合中的数据生产LinkedList
    }

boolean addAll(Collection<?  c) {
        return addAll(size,c);
    }

boolean addAll(int index,Collection<?  c) {
        checkPositionIndex(index);

        Object[] a = c.toArray(); //把集合转为数组
        int numNew = a.length;
        if (numNew == 0)
            return false;

        Node<E> pred,succ;
        if (index == size) {
            succ = null;
            pred = last;
        } else {
            succ = node(index);
            pred = succ.prev;
        }

        for (Object o : a) { //对数组进行遍历,对每一个元素都封装成Node并添加到LinkedList中
            @SuppressWarnings("unchecked") E e = (E) o;
            Node<E> newNode = new Node<>(pred,e,);
            if (pred == )
                first = newNode;
            
                pred.next = newNode;
            pred = newNode;
        }

        if (succ == ) {
            last = pred;
        }  {
            pred.next = succ;
            succ.prev = pred;
        }

        size += numNew;
        modCount++;
        true;
    }

接下来看看LinkedList的基本操作,添加,删除,遍历,查询等

先看添加,从双向链表的结构来看,添加元素可以在链表的头、尾、以及中间的任意位置添加新的元素。因为 LinkedList 有头指针和尾指针,所以在表头或表尾进 行插入元素只需要 O(1) 的时间,而在指定位置插入元素则需要先遍历一下链表, 所以复杂度为 O(n)。首先看看在头部添加元素:

 

 

 看图可以看出,只要把first指向新的node,新的node的next指向原先firt指向的node,再把原先first指向的node的prev指向新的node就可以了。


     * Links e as first element.
     void linkFirst(E e) {
        final Node<E> f = first; //使用临时node
        final Node<E> newNode = new Node<>( newNode;
        if (f == ) //判断first是否为空
            last =
            f.prev = newNode; //把f的prev指向新的node
        size++; //链表长度加1
        modCount++; //记录链表被修改的次数
    }

在看看在尾部添加,其实和在头部添加一样,只是把first换成了last,逻辑一样


     * Links e as last element.
      linkLast(E e) {
        final Node<E> l = last;
        new Node<>(l,1)">);
        last =if (l == )
            first =
            l.next = newNode;
        size++;
        modCount++;
    }

再看看在中间的任意位置添加:

 

 

 这个相对来说复杂点点,修改添加前后node的next和prev的指向,修改的相对来说多点点


     * Inserts element e before non-null Node succ.
     void linkBefore(E e,1)"> succ) { //表示在在succ节点前面添加e元素
        // assert succ != null;
        final Node<E> pred = succ.prev; //获取succ的前面节点
        new Node<>(pred,succ); //把e封装成节点,并把prev指向succ前面节点,把next指向succ节点
        succ.prev = newNode; //然后把succ的prev指向新的节点
        
            pred.next = newNode; //把succ的前节点的next只想新的节点
        size++; //链表长度+1
        modCount++; //修改次数+1
    }

添加说完了,就说说删除,其实也很简单

 

 

 删除也是分为从头部、尾部、中间位置删除

先看看从first位置删除


     * Unlinks non-null first node f. 
     private E unlinkFirst(Node<E> f) { 
         assert f == first && f != null; 
        final E element = f.item; //获取first中间的元素,用于后面的返回
        final Node<E> next = f.next; //获取f的next节点
        f.item = ;
        f.next = null;  help GC 清除
        first = next; //把first指向f的next
        if (next == )
            last = 
            next.prev = ;  //清除
        size--; //链表长度-1
        modCount++; //修改次数+1
         element;
    }

看了从头部删除,其实尾部删除也差不多

 
     * Unlinks non-null last node l.
     private E unlinkLast(Node<E> l) {
         assert l == last && l != null;
         l.item;
        final Node<E> prev = l.prev;
        l.item = ;
        l.prev =  help GC
        last = prev;
        if (prev == )
            first = 
            prev.next = ;
        size-- element;
    }

在看看从指定位置删除吧


     * Unlinks non-null node x.
     */
    E unlink(Node<E> x) {
         assert x != null;
         x.item; //获取该节点的值
         x.next; //获取该节点的next节点
         x.prev; //获取该节点的prev节点

        ) { //把该节点的前节点的next指向该节点的next节点,并清除该节点的prev指向
            first = next;
        }  {
            prev.next = next;
            x.prev = ;
        }

        ) { //把该节点的next节点的prev指向该节点的prev节点,并清除该节点的next指向
            last = prev;
        }  {
            next.prev = prev;
            x.next = ;
        }

        x.item = ; //清除
        size-- element;
    }

看完增删,那就继续看查相关的方法,也有从头,尾相关的查询方法,都很简单,做判断,然后查询


     * Returns the first element in this list.
     *
     * @return the first element in this list
     *  NoSuchElementException if this list is empty
      E getFirst() {
         first;
        throw new NoSuchElementException();
         f.item;
    }

    
     * Returns the last element in this list.
     *
     *  the last element in this list
     *  E getLast() {
         l.item;
    }

当然还有指定index查询的


     * Returns the (non-null) Node at the specified element index.
     
    Node<E> node(int index) {
         assert isElementIndex(index);
        //判断index是在链表的前半段还是在后半段,如果在前半段就从first向后遍历,否则使用last向前遍历
        if (index < (size >> 1)) { 
            Node<E> x = first;
            for (int i = 0; i < index; i++)
                x = x.next;
             x;
        }  {
            Node<E> x = last;
            int i = size - 1; i > index; i-- x.prev;
             x;
        }
    }

其实基本知道了上面的方法基本对双向链表有了一定的熟悉,当然LinkedList还有很多其他的方法,不过很多都是基于上面这些方法的一些封装,例如:


     * Inserts the specified element at the beginning of this list.
     *
     *  e the element to add
      addFirst(E e) {
        linkFirst(e);
    }

    
     * Appends the specified element to the end of this list.
     *
     * <p>This method is equivalent to {@link #add}.
     *
     *  addLast(E e) {
        linkLast(e);
    }

     * Removes and returns the first element from this list.
     *
     *  the first element from this list
     *  E removeFirst() {
         unlinkFirst(f);
    }

    
     * Removes and returns the last element from this list.
     *
     *  the last element from this list
     *  E removeLast() {
         unlinkLast(l);
    }
 #addLast}.
     *
     *  e element to be appended to this list
     *  {@code true} (as specified by { Collection#add})
     boolean add(E e) {
        linkLast(e);
        ;
    }

     * Removes the first occurrence of the specified element from this list,* if it is present.  If this list does not contain the element,it is
     * unchanged.  More formally,removes the element with the lowest index
     * { i} such that
     * <tt>(o==null&nbsp;?&nbsp;get(i)==null&nbsp;:&nbsp;o.equals(get(i)))</tt>
     * (if such an element exists).  Returns { true} if this list
     * contained the specified element (or equivalently,if this list
     * changed as a result of the call).
     *
     *  o element to be removed from this list,if present
     *  true} if this list contained the specified element
      remove(Object o) {
        if (o == ) {
            for (Node<E> x = first; x != null; x = x.next) {
                if (x.item == ) {
                    unlink(x);
                    ;
                }
            }
        }  {
            if (o.equals(x.item)) {
                    unlink(x);
                    ;
                }
            }
        }
        
     * Removes all of the elements from this list.
     * The list will be empty after this call returns.
      clear() {
         Clearing all of the links between nodes is "unnecessary",but:
         - helps a generational GC if the discarded nodes inhabit
           more than one generation
         - is sure to free memory even if there is a reachable Iterator
        ; ) {
            Node<E> next = x.next;
            x.item = ;
            x.next = ;
            x.prev = ;
            x = next;
        }
        first = last = ;
        size = 0
     * Removes the element at the specified position in this list.  Shifts any
     * subsequent elements to the left (subtracts one from their indices).
     * Returns the element that was removed from the list.
     *
     *  index the index of the element to be removed
     *  the element previously at the specified position
     *  IndexOutOfBoundsException {@inheritDoc}
     public E remove( index) {
        checkElementIndex(index);
         unlink(node(index));
    }

     * Returns the index of the first occurrence of the specified element
     * in this list,or -1 if this list does not contain the element.
     * More formally,returns the lowest index { i} such that
     * <tt>(o==null&nbsp;?&nbsp;get(i)==null&nbsp;:&nbsp;o.equals(get(i)))</tt>,* or -1 if there is no such index.
     *
     *  o element to search for
     *  the index of the first occurrence of the specified element in
     *         this list,or -1 if this list does not contain the element
      indexOf(Object o) { //查找元素o是否在链表中,并返回index,没找到返回-1
        int index = 0)
                     index;
                index++;
            }
        }  (o.equals(x.item))
                    ;
            }
        }
        return -1;
    }

到这里本文就结束了了,如果想知道LinkedList的更多方法,建议去看源码

相关文章

摘要: 原创出处 https://www.bysocket.com 「公众号:泥瓦匠...
摘要: 原创出处 https://www.bysocket.com 「公众号:泥瓦匠...
今天犯了个错:“接口变动,伤筋动骨,除非你确定只有你一个...
Writer :BYSocket(泥沙砖瓦浆木匠)微 博:BYSocket豆 瓣:...
本文目录 线程与多线程 线程的运行与创建 线程的状态 1 线程...