zoukankan      html  css  js  c++  java
  • 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

    作者:一寸HUI
    出处:https://www.cnblogs.com/zsql/
    如果您觉得阅读本文对您有帮助,请点击一下右下方的推荐按钮,您的推荐将是我写作的最大动力!
    版权声明:本文为博主原创或转载文章,欢迎转载,但转载文章之后必须在文章页面明显位置注明出处,否则保留追究法律责任的权利。
  • 相关阅读:
    HDU2149-Public Sale
    分页和多条件查询功能
    hdu 4691 最长的共同前缀 后缀数组 +lcp+rmq
    BZOJ 2588 Count on a tree (COT) 是持久的段树
    windows 设置脚本IP
    hdu 4912 Paths on the tree(树链拆分+贪婪)
    分散式-ubuntu12.04安装hadoop1.2.1
    struts详细解释拦截器
    Codeforces 459E Pashmak and Graph(dp+贪婪)
    C#中的数据格式转换 (未完待更新)
  • 原文地址:https://www.cnblogs.com/zsql/p/14412409.html
Copyright © 2011-2022 走看看