Collection - LinkedList源码解析
约 3935 字大约 13 分钟
Collection - LinkedList源码解析
本文主要对Collection - LinkedList进行源码解析。@pdai
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 ? get(i)==null : 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; // GC
size--;
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 GC
first = next;
if (next == null)
last = null;
else
next.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 GC
last = prev;
if (prev == null)
first = null;
else
prev.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;
else
l.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);
else
linkBefore(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;
else
pred.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 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++;
}
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);
else
linkBefore(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 ? get(i)==null : 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 ? get(i)==null : 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;
}
参考
- Java LinkedList源码剖析 结合源码对LinkedList进行讲解 http://www.cnblogs.com/CarpenterLee/p/5457150.html