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。
链表这种数据结构本身的原理（单链表、双向链表的插入删除过程）可以参考算法 003：线性表：链表，本文聚焦 LinkedList 的源码实现。
| 操作 | 方法 | 时间复杂度 |
|---|---|---|
| 首尾插入/删除 | addFirst(e) / addLast(e) / removeFirst() / removeLast() | O(1) |
| 按下标访问/替换 | get(index) / set(index, e) | O(n)，node(index) 从较近的一端折半查找 |
| 按下标插入/删除 | add(index, e) / remove(index) | O(n)，需先线性定位，再 O(1) 修改引用 |
| 按值查找/删除 | indexOf(o) / contains(o) / remove(o) | O(n) |
LinkedList 的实现
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，也即是先解跟 List 接口相关的函数，而将 Queue、Stack 以及 Deque 相关的知识放在后文讲。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。
flowchart LR
first["first"] --> A
subgraph nodes [" "]
A["Node\nitem = A\nprev = null"] -->|next| B["Node\nitem = B"]
B -->|prev| A
B -->|next| C["Node\nitem = C\nnext = null"]
C -->|prev| B
end
C --> last["last"]（原文配图已丢失，据源码与文字描述重绘：每个 Node 含 item、next、prev 三个域；空链表时 first 与 last 均为 null。）
其中 Node 是私有的内部类：
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;
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.
*
* @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)，该方法是在指定下标处插入元素，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));
}
/**
* 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;
}
}上面代码中的 node(int index) 函数有一点小小的 trick：因为链表是双向的，可以从开始往后找，也可以从结尾往前找，具体朝哪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。
addAll()
addAll(index, c) 实现方式并不是直接调用 add(index, e) 来实现，主要是效率的问题，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.
*
* @param c collection whose elements are 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.
*
* @param index index at which to insert the first element from the
* specified collection
* @param c collection whose 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、删除指定位置的元素，即前文的 add(int index, E element) 和 remove(int index)，不再重复。下标合法性检查由下面几个辅助方法完成：
/**
* 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；查找最后一次出现的 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.
*/
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;
}
/**
* Returns 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}
* @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}
* @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}
* @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}
* @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).
*
* @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).
*
* @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;
}系列导航
- 上一篇：ArrayList 源码解析
- 下一篇：Stack & Queue 源码解析