java集合之LinkedList分析
LinkedList概述
LinkedList继承自AbstractSequentialList，实现了List接口和Deque接口。
既可以当做list用，也可作为队列和栈来用。可以说是非常的全面了。
本文分析基于jdk1.8
LinkedList底层实现原理
底层数据结构
LinkedList底层采用双向链表实现。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。
在这里没有所谓的哑元，当链表为空的时候first和last都指向null。
Node的数据结构：
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;
}
}
first和last节点:
/**
* 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;
时间复杂度
LinkedList的实现方式决定了所有跟下表有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的添加操作
add(E e)方法
add()方法有两个版本，一个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，只需简单修改几个相关引用即可；
另外一个add(int index,E e),该方法是在指定下标出插入元素，需要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。
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++;
}
/**
* Links e as first element.
*/
private void linkFirst(E e) {
final Node<E> f = first;
final Node<E> newNode = new Node<>(null, e, f);
first = newNode;
if (f == null)
last = newNode;
else
f.prev = newNode;
size++;
modCount++;
}
add(int index, E element)方法
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trick,因为链表时双向的，可以从后往前找，也可以从前往后招，具体朝哪个方向取决于条件 index < (size >> 1) ，也就是index是靠近前端还是后端。从这里可以看出，linkedList通过index检查元素的效率没有arrayList高。
addAll(Collection<? extends E> c)方法
addAll(index,c)实现方式并不是直接调用add(index,e)实现的，主要是因为效率问题，另一个就是fail-fast中modCount只会增加1次；
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();
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;
}
addFirst(E e)和addLast(E e)方法
/**
* 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);
}
基于队列接口的offer方法
public boolean offer(E e) {
return add(e);
}
// Deque operations
/**
* 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;
}
基于栈的push方法
public void push(E e) {
addFirst(e);
}
LinkedList的删除操作
remove()和removeFirst()方法
/**
* 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();
}
/**
* 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;
}
基于队列的poll()方法
public E poll() {
final Node<E> f = first;
return (f == null) ? null : unlinkFirst(f);
}
基于栈的pop()方法
/**
* 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();
}
LinkedList的查找操作
LinkedList 底层基于链表结构，无法向 ArrayList 那样随机访问指定位置的元素。LinkedList 查找过程要稍麻烦一些，需要从链表头结点（或尾节点）向后查找，时间复杂度为 O(n)。
get(int index)方法
/**
* 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;
}
/**
* 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;
}
}
private void checkElementIndex(int index) {
if (!isElementIndex(index))
throw new IndexOutOfBoundsException(outOfBoundsMsg(index));
}
private boolean isElementIndex(int index) {
return index >= 0 && index < size;
}
主要是通过遍历的方式定位目标位置的节点。获取到节点后，取出节点存储的值返回即可。这里面有个小优化，即通过比较 index 与节点数量 size/2 的大小，决定从头结点还是尾节点进行查找。
LinkedList的遍历操作
链表的遍历过程也很简单，和上面查找过程类似，我们从头节点往后遍历就行了。但对于 LinkedList 的遍历还是需要注意一些，不然可能会导致代码效率低下。通常情况下，我们会使用 foreach 遍历 LinkedList，而 foreach 最终转换成迭代器形式。所以分析 LinkedList 的遍历的核心就是它的迭代器实现。
public ListIterator<E> listIterator(int index) {
checkPositionIndex(index);
return new ListItr(index);
}
private class ListItr implements ListIterator<E> {
private Node<E> lastReturned;
private Node<E> next;
private int nextIndex;
private int expectedModCount = modCount;
ListItr(int index) {
// assert isPositionIndex(index);
next = (index == size) ? null : node(index);
nextIndex = index;
}
public boolean hasNext() {
return nextIndex < size;
}
public E next() {
checkForComodification();
if (!hasNext())
throw new NoSuchElementException();
lastReturned = next;
next = next.next;
nextIndex++;
return lastReturned.item;
}
public boolean hasPrevious() {
return nextIndex > 0;
}
public E previous() {
checkForComodification();
if (!hasPrevious())
throw new NoSuchElementException();
lastReturned = next = (next == null) ? last : next.prev;
nextIndex--;
return lastReturned.item;
}
public int nextIndex() {
return nextIndex;
}
public int previousIndex() {
return nextIndex - 1;
}
public void remove() {
checkForComodification();
if (lastReturned == null)
throw new IllegalStateException();
Node<E> lastNext = lastReturned.next;
unlink(lastReturned);
if (next == lastReturned)
next = lastNext;
else
nextIndex--;
lastReturned = null;
expectedModCount++;
}
public void set(E e) {
if (lastReturned == null)
throw new IllegalStateException();
checkForComodification();
lastReturned.item = e;
}
public void add(E e) {
checkForComodification();
lastReturned = null;
if (next == null)
linkLast(e);
else
linkBefore(e, next);
nextIndex++;
expectedModCount++;
}
public void forEachRemaining(Consumer<? super E> action) {
Objects.requireNonNull(action);
while (modCount == expectedModCount && nextIndex < size) {
action.accept(next.item);
lastReturned = next;
next = next.next;
nextIndex++;
}
checkForComodification();
}
final void checkForComodification() {
if (modCount != expectedModCount)
throw new ConcurrentModificationException();
}
}
LinkedList不擅长随机位置访问，如果使用随机访问遍历LinkedList效率会很差，如下：
List<Integer> list = new LinkedList<>();
list.add(1);
list.add(2);
for (int i = 0; i < list.size(); i++) {
int t = list.get(i);
}
