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 } public boolean addAll(Collection<? extends E> c) { return addAll(size, c); } public boolean addAll(int index, Collection<? extends E> 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, null); if (pred == null) first = newNode; else pred.next = newNode; pred = newNode; } if (succ == null) { last = pred; } else { pred.next = succ; succ.prev = pred; } size += numNew; modCount++; return 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. */
    private void linkFirst(E e) { final Node<E> f = first; //使用临时node final Node<E> newNode = new Node<>(null, e, f); //封装新的node,并把新node的nex指向f first = newNode; if (f == null) //判断first是否为空 last = newNode; else 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. */
    void linkLast(E e) { final Node<E> l = last; final Node<E> newNode = new Node<>(l, e, null); last = newNode; if (l == null) first = newNode; else 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, Node<E> 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 final Node<E> newNode = new Node<>(pred, e, 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 if (pred == null) first = newNode; else 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 = null; f.next = null; // help GC 清除
        first = next; //把first指向f的next if (next == null) last = null; else next.prev = null; //清除 size--; //链表长度-1 modCount++; //修改次数+1 return 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;
        final E element = l.item; final Node<E> prev = l.prev; l.item = null; l.prev = null; // help GC
        last = prev; if (prev == null) first = null; else prev.next = null; size--; modCount++; return 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;
        final E element = x.item; //获取该节点的值 final Node<E> next = x.next; //获取该节点的next节点 final Node<E> prev = x.prev; //获取该节点的prev节点 if (prev == null) { //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; } else { prev.next = next; x.prev = null; } if (next == null) { //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; } else { next.prev = prev; x.next = null; } x.item = null; //清除 size--; //链表长度-1 modCount++; //修改次数+1 return 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 * @throws NoSuchElementException if this list is empty */
    public E getFirst() { final Node<E> f = first; if (f == null) throw new NoSuchElementException(); return f.item; } /** * Returns the last element in this list. * * @return the last element in this list * @throws NoSuchElementException if this list is empty */
    public E getLast() { final Node<E> l = last; if (l == null) throw new NoSuchElementException(); return 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; return x; } else { Node<E> x = last; for (int i = size - 1; i > index; i--) x = x.prev; return 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. * * @param e the element to add */
    public void addFirst(E e) { linkFirst(e); } /** * Appends the specified element to the end of this list. * * <p>This method is equivalent to {@link #add}. * * @param e the element to add */
    public void addLast(E e) { linkLast(e); } /** * Removes and returns the first element from this list. * * @return the first element from this list * @throws NoSuchElementException if this list is empty */
    public E removeFirst() { final Node<E> f = first; if (f == null) throw new NoSuchElementException(); return unlinkFirst(f); } /** * Removes and returns the last element from this list. * * @return the last element from this list * @throws NoSuchElementException if this list is empty */
    public E removeLast() { final Node<E> l = last; if (l == null) throw new NoSuchElementException(); return unlinkLast(l); } /** * Appends the specified element to the end of this list. * * <p>This method is equivalent to {@link #addLast}. * * @param e element to be appended to this list * @return {@code true} (as specified by {@link Collection#add}) */
    public boolean add(E e) { linkLast(e); return true; } /** * 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 * {@code i} such that * <tt>(o==null&nbsp;?&nbsp;get(i)==null&nbsp;:&nbsp;o.equals(get(i)))</tt> * (if such an element exists). Returns {@code true} if this list * contained the specified element (or equivalently, if this list * changed as a result of the call). * * @param o element to be removed from this list, if present * @return {@code true} if this list contained the specified element */
    public boolean remove(Object o) { if (o == null) { for (Node<E> x = first; x != null; x = x.next) { if (x.item == null) { unlink(x); return true; } } } else { for (Node<E> x = first; x != null; x = x.next) { if (o.equals(x.item)) { unlink(x); return true; } } } return false; } /** * Removes all of the elements from this list. * The list will be empty after this call returns. */
    public void 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
        for (Node<E> x = first; x != null; ) { Node<E> next = x.next; x.item = null; x.next = null; x.prev = null; x = next; } first = last = null; size = 0; modCount++; } /** * 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. * * @param index the index of the element to be removed * @return the element previously at the specified position * @throws IndexOutOfBoundsException {@inheritDoc} */
    public E remove(int index) { checkElementIndex(index); return 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 {@code 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. * * @param o element to search for * @return the index of the first occurrence of the specified element in * this list, or -1 if this list does not contain the element */
    public int 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; if (o == null) { for (Node<E> x = first; x != null; x = x.next) { if (x.item == null) return index; index++; } } else { for (Node<E> x = first; x != null; x = x.next) { if (o.equals(x.item)) return index; index++; } } 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

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