【Collection - LinkedList源码解析】

本文主要对Collection - LinkedList进行源码解析。

  • Collection - LinkedList源码解析
    • 概述
    • LinkedList实现
      • 底层数据结构
      • 构造函数
      • getFirst(), getLast()
      • removeFirst(), removeLast(), remove(e), remove(index)
      • add()
      • addAll()
      • clear()
      • Positional Access 方法
      • 查找操作
      • Queue 方法
      • Deque 方法

 概述

LinkedList同时实现了List接口和Deque接口，也就是说它既可以看作一个顺序容器，又可以看作一个队列(Queue)，同时又可以看作一个栈(Stack)。这样看来，LinkedList简直就是个全能冠军。当你需要使用栈或者队列时，可以考虑使用LinkedList，一方面是因为Java官方已经声明不建议使用Stack类，更遗憾的是，Java里根本没有一个叫做Queue的类(它是个接口名字)。关于栈或队列，现在的首选是ArrayDeque，它有着比LinkedList(当作栈或队列使用时)有着更好的性能。

LinkedList的实现方式决定了所有跟下标相关的操作都是线性时间，而在首段或者末尾删除元素只需要常数时间。为追求效率LinkedList没有实现同步(synchronized)，如果需要多个线程并发访问，可以先采用Collections.synchronizedList()方法对其进行包装。

 LinkedList实现

 底层数据结构

LinkedList底层通过双向链表实现，本节将着重讲解插入和删除元素时双向链表的维护过程，也即是之间解跟List接口相关的函数，而将Queue和Stack以及Deque相关的知识放在下一节讲。双向链表的每个节点用内部类Node表示。LinkedList通过first和last引用分别指向链表的第一个和最后一个元素。注意这里没有所谓的哑元，当链表为空的时候first和last都指向null。

    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;

其中Node是私有的内部类:

    private static class Node<E> {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;}}

 构造函数

    /*** Constructs an empty list.*/public LinkedList() {}/*** 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) {this();addAll(c);}

 getFirst(), getLast()

获取第一个元素， 和获取最后一个元素:

    /*** 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;}

 removeFirst(), removeLast(), remove(e), remove(index)

remove()方法也有两个版本，一个是删除跟指定元素相等的第一个元素remove(Object o)，另一个是删除指定下标处的元素remove(int index)。

删除元素 - 指的是删除第一次出现的这个元素, 如果没有这个元素，则返回false；判断的依据是equals方法， 如果equals，则直接unlink这个node；由于LinkedList可存放null元素，故也可以删除第一次出现null的元素；

    /*** 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;}/*** Unlinks non-null node x.*/E unlink(Node<E> x) {// assert x != null;final E element = x.item;final Node<E> next = x.next;final Node<E> prev = x.prev;if (prev == null) {// 第一个元素first = next;} else {prev.next = next;x.prev = null;}if (next == null) {// 最后一个元素last = prev;} else {next.prev = prev;x.next = null;}x.item = null; // GCsize--;modCount++;return element;}

remove(int index)使用的是下标计数， 只需要判断该index是否有元素即可，如果有则直接unlink这个node。

    /*** 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));}

删除head元素:

    /*** 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);}/*** Unlinks non-null first node f.*/private E unlinkFirst(Node<E> f) {// assert f == first && f != null;final E element = f.item;final Node<E> next = f.next;f.item = null;f.next = null; // help GCfirst = next;if (next == null)last = null;elsenext.prev = null;size--;modCount++;return element;}

删除last元素:

	/*** 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);}/*** 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 GClast = prev;if (prev == null)first = null;elseprev.next = null;size--;modCount++;return element;}

 add()

add()方法有两个版本，一个是add(E e)，该方法在LinkedList的末尾插入元素，因为有last指向链表末尾，在末尾插入元素的花费是常数时间。只需要简单修改几个相关引用即可；另一个是add(int index, E element)，该方法是在指定下表处插入元素，需要先通过线性查找找到具体位置，然后修改相关引用完成插入操作。

    /*** 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;}/*** 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;elsel.next = newNode;size++;modCount++;}

add(int index, E element), 当index==size时，等同于add(E e); 如果不是，则分两步: 1.先根据index找到要插入的位置,即node(index)方法；2.修改引用，完成插入操作。

    /*** Inserts the specified element at the specified position in this list.* Shifts the element currently at that position (if any) and any* subsequent elements to the right (adds one to their indices).** @param index index at which the specified element is to be inserted* @param element element to be inserted* @throws IndexOutOfBoundsException {@inheritDoc}*/public void add(int index, E element) {checkPositionIndex(index);if (index == size)linkLast(element);elselinkBefore(element, node(index));}

上面代码中的node(int index)函数有一点小小的trick，因为链表双向的，可以从开始往后找，也可以从结尾往前找，具体朝那个方向找取决于条件index < (size >> 1)，也即是index是靠近前端还是后端。从这里也可以看出，linkedList通过index检索元素的效率没有arrayList高。

    /*** Returns the (non-null) Node at the specified element index.*/Node<E> node(int index) {// assert isElementIndex(index);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;}}

 addAll()

addAll(index, c) 实现方式并不是直接调用add(index,e)来实现，主要是因为效率的问题，另一个是fail-fast中modCount只会增加1次；

    /*** Appends all of the elements in the specified collection to the end of* this list, in the order that they are returned by the specified* collection's iterator.  The behavior of this operation is undefined if* the specified collection is modified while the operation is in* progress.  (Note that this will occur if the specified collection is* this list, and it's nonempty.)** @param c collection containing elements to be added to this list* @return {@code true} if this list changed as a result of the call* @throws NullPointerException if the specified collection is null*/public boolean addAll(Collection<? extends E> c) {return addAll(size, c);}/*** Inserts all of the elements in the specified collection into this* list, starting at the specified position.  Shifts the element* currently at that position (if any) and any subsequent elements to* the right (increases their indices).  The new elements will appear* in the list in the order that they are returned by the* specified collection's iterator.** @param index index at which to insert the first element*              from the specified collection* @param c collection containing elements to be added to this list* @return {@code true} if this list changed as a result of the call* @throws IndexOutOfBoundsException {@inheritDoc}* @throws NullPointerException if the specified collection is null*/public boolean addAll(int index, Collection<? extends E> c) {checkPositionIndex(index);Object[] a = c.toArray();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) {@SuppressWarnings("unchecked") E e = (E) o;Node<E> newNode = new Node<>(pred, e, null);if (pred == null)first = newNode;elsepred.next = newNode;pred = newNode;}if (succ == null) {last = pred;} else {pred.next = succ;succ.prev = pred;}size += numNew;modCount++;return true;}

 clear()

为了让GC更快可以回收放置的元素，需要将node之间的引用关系赋空。

    /*** 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 Iteratorfor (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++;}

 Positional Access 方法

通过index获取元素

    /*** Returns the element at the specified position in this list.** @param index index of the element to return* @return the element at the specified position in this list* @throws IndexOutOfBoundsException {@inheritDoc}*/public E get(int index) {checkElementIndex(index);return node(index).item;}

将某个位置的元素重新赋值:

    /*** Replaces the element at the specified position in this list with the* specified element.** @param index index of the element to replace* @param element element to be stored at the specified position* @return the element previously at the specified position* @throws IndexOutOfBoundsException {@inheritDoc}*/public E set(int index, E element) {checkElementIndex(index);Node<E> x = node(index);E oldVal = x.item;x.item = element;return oldVal;}

将元素插入到指定index位置:

    /*** Inserts the specified element at the specified position in this list.* Shifts the element currently at that position (if any) and any* subsequent elements to the right (adds one to their indices).** @param index index at which the specified element is to be inserted* @param element element to be inserted* @throws IndexOutOfBoundsException {@inheritDoc}*/public void add(int index, E element) {checkPositionIndex(index);if (index == size)linkLast(element);elselinkBefore(element, node(index));}

删除指定位置的元素:

    /*** 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));}

其它位置的方法:

/*** Tells if the argument is the index of an existing element.*/private boolean isElementIndex(int index) {return index >= 0 && index < size;}/*** Tells if the argument is the index of a valid position for an* iterator or an add operation.*/private boolean isPositionIndex(int index) {return index >= 0 && index <= size;}/*** Constructs an IndexOutOfBoundsException detail message.* Of the many possible refactorings of the error handling code,* this "outlining" performs best with both server and client VMs.*/private String outOfBoundsMsg(int index) {return "Index: "+index+", Size: "+size;}private void checkElementIndex(int index) {if (!isElementIndex(index))throw new IndexOutOfBoundsException(outOfBoundsMsg(index));}private void checkPositionIndex(int index) {if (!isPositionIndex(index))throw new IndexOutOfBoundsException(outOfBoundsMsg(index));}

 查找操作

查找操作的本质是查找元素的下标:

查找第一次出现的index, 如果找不到返回-1；

    /*** 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) {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;}

查找最后一次出现的index, 如果找不到返回-1；

    /*** Returns the index of the last occurrence of the specified element* in this list, or -1 if this list does not contain the element.* More formally, returns the highest 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 last occurrence of the specified element in*         this list, or -1 if this list does not contain the element*/public int lastIndexOf(Object o) {int index = size;if (o == null) {for (Node<E> x = last; x != null; x = x.prev) {index--;if (x.item == null)return index;}} else {for (Node<E> x = last; x != null; x = x.prev) {index--;if (o.equals(x.item))return index;}}return -1;}

 Queue 方法

   /*** Retrieves, but does not remove, the head (first element) of this list.** @return the head of this list, or {@code null} if this list is empty* @since 1.5*/public E peek() {final Node<E> f = first;return (f == null) ? null : f.item;}/*** Retrieves, but does not remove, the head (first element) of this list.** @return the head of this list* @throws NoSuchElementException if this list is empty* @since 1.5*/public E element() {return getFirst();}/*** Retrieves and removes the head (first element) of this list.** @return the head of this list, or {@code null} if this list is empty* @since 1.5*/public E poll() {final Node<E> f = first;return (f == null) ? null : unlinkFirst(f);}/*** Retrieves and removes the head (first element) of this list.** @return the head of this list* @throws NoSuchElementException if this list is empty* @since 1.5*/public E remove() {return removeFirst();}/*** Adds the specified element as the tail (last element) of this list.** @param e the element to add* @return {@code true} (as specified by {@link Queue#offer})* @since 1.5*/public boolean offer(E e) {return add(e);}

 Deque 方法

    /*** Inserts the specified element at the front of this list.** @param e the element to insert* @return {@code true} (as specified by {@link Deque#offerFirst})* @since 1.6*/public boolean offerFirst(E e) {addFirst(e);return true;}/*** Inserts the specified element at the end of this list.** @param e the element to insert* @return {@code true} (as specified by {@link Deque#offerLast})* @since 1.6*/public boolean offerLast(E e) {addLast(e);return true;}/*** Retrieves, but does not remove, the first element of this list,* or returns {@code null} if this list is empty.** @return the first element of this list, or {@code null}*         if this list is empty* @since 1.6*/public E peekFirst() {final Node<E> f = first;return (f == null) ? null : f.item;}/*** Retrieves, but does not remove, the last element of this list,* or returns {@code null} if this list is empty.** @return the last element of this list, or {@code null}*         if this list is empty* @since 1.6*/public E peekLast() {final Node<E> l = last;return (l == null) ? null : l.item;}/*** Retrieves and removes the first element of this list,* or returns {@code null} if this list is empty.** @return the first element of this list, or {@code null} if*     this list is empty* @since 1.6*/public E pollFirst() {final Node<E> f = first;return (f == null) ? null : unlinkFirst(f);}/*** Retrieves and removes the last element of this list,* or returns {@code null} if this list is empty.** @return the last element of this list, or {@code null} if*     this list is empty* @since 1.6*/public E pollLast() {final Node<E> l = last;return (l == null) ? null : unlinkLast(l);}/*** Pushes an element onto the stack represented by this list.  In other* words, inserts the element at the front of this list.** <p>This method is equivalent to {@link #addFirst}.** @param e the element to push* @since 1.6*/public void push(E e) {addFirst(e);}/*** Pops an element from the stack represented by this list.  In other* words, removes and returns the first element of this list.** <p>This method is equivalent to {@link #removeFirst()}.** @return the element at the front of this list (which is the top*         of the stack represented by this list)* @throws NoSuchElementException if this list is empty* @since 1.6*/public E pop() {return removeFirst();}/*** Removes the first occurrence of the specified element in this* list (when traversing the list from head to tail).  If the list* does not contain the element, it is unchanged.** @param o element to be removed from this list, if present* @return {@code true} if the list contained the specified element* @since 1.6*/public boolean removeFirstOccurrence(Object o) {return remove(o);}/*** Removes the last occurrence of the specified element in this* list (when traversing the list from head to tail).  If the list* does not contain the element, it is unchanged.** @param o element to be removed from this list, if present* @return {@code true} if the list contained the specified element* @since 1.6*/public boolean removeLastOccurrence(Object o) {if (o == null) {for (Node<E> x = last; x != null; x = x.prev) {if (x.item == null) {unlink(x);return true;}}} else {for (Node<E> x = last; x != null; x = x.prev) {if (o.equals(x.item)) {unlink(x);return true;}}}return false;}

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