文章目錄
Pre
概述
從上圖可知: LinkedList同時(shí)實(shí)現(xiàn)了List接口和Deque接口,既可以看作一個(gè)順序容器,又可以看作一個(gè)隊(duì)列(Queue),同時(shí)又可以看作一個(gè)棧(Stack) 。
當(dāng)需要使用?;蛘哧?duì)列時(shí),可以考慮使用LinkedList,一方面是因?yàn)镴ava官方已經(jīng)聲明不建議使用Stack類,更遺憾的是,Java里根本沒有一個(gè)叫做Queue的類(它是個(gè)接口名字)。
關(guān)于?;蜿?duì)列,現(xiàn)在的首選是ArrayDeque,它有著比LinkedList(當(dāng)作棧或隊(duì)列使用時(shí))有著更好的性能。
ArrayDeque 移步 https://www.nhooo.com/java/java-arraydeque.html
LinkedList的實(shí)現(xiàn)方式?jīng)Q定了所有跟下標(biāo)相關(guān)的操作都是線性時(shí)間,而在首段或者末尾刪除元素只需要常數(shù)時(shí)間。
為追求效率LinkedList沒有實(shí)現(xiàn)同步(synchronized),如果需要多個(gè)線程并發(fā)訪問,可以先采用Collections.synchronizedList()方法對(duì)其進(jìn)行包裝。
底層數(shù)據(jù)結(jié)構(gòu)-雙向鏈表
-
LinkedList底層通過雙向鏈表實(shí)現(xiàn) 。
-
雙向鏈表的每個(gè)節(jié)點(diǎn)用內(nèi)部類Node表示。LinkedList通過first和last引用分別指向鏈表的第一個(gè)和最后一個(gè)元素。
-
沒有所謂的啞元,當(dāng)鏈表為空的時(shí)候first和last都指向null。
其中Node是私有的內(nèi)部類:
源碼解析
構(gòu)造函數(shù)
/**
* 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);
}
方法源碼分析
getFirst()
獲取第一個(gè)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;
}
getLast()
獲取最后一個(gè)元素
/**
* 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;
}
remove相關(guān)方法
有兩個(gè)版本,
- 一個(gè)是刪除跟指定元素相等的第一個(gè)元素
remove(Object o)
, - 一個(gè)是刪除指定下標(biāo)處的元素
remove(int index)
remove(e)
【boolean remove(Object o) 】
/**
* 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) { //刪除的是第一個(gè)元素
first = next;
} else {
prev.next = next;
x.prev = null;
}
if (next == null) {//刪除的是最后一個(gè)元素
last = prev;
} else {
next.prev = prev;
x.next = null;
}
x.item = null;
size--;
modCount++;
return element;
}
指的是刪除第一次出現(xiàn)的這個(gè)元素, 如果沒有這個(gè)元素,則返回false;判讀的依據(jù)是equals方法, 如果equals,則直接unlink這個(gè)node;由于LinkedList可存放null元素,故也可以刪除第一次出現(xiàn)null的元素;
remove(index)
【remove(int index)】
remove(int index)使用的是下標(biāo)計(jì)數(shù), 只需要判斷該index是否有元素即可,如果有則直接unlink這個(gè)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));
}
removeFirst()
【刪除 頭部元素】
/**
* 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;
}
removeLast()
/**
* 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()方法有兩個(gè)版本:
- 一個(gè)是
add(E e)
,該方法在LinkedList的末尾插入元素,因?yàn)橛衛(wèi)ast指向鏈表末尾,在末尾插入元素的花費(fèi)是常數(shù)時(shí)間。只需要簡(jiǎn)單修改幾個(gè)相關(guān)引用即可 - 另一個(gè)是
add(int index, E element)
,該方法是在指定下表處插入元素,需要先通過線性查找找到具體位置,然后修改相關(guān)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),
- 當(dāng)index==size時(shí),等同于add(E e);
- 如果不是,則分兩步: 1.先根據(jù)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));
}
/**
* Inserts element e before non-null Node succ.
*/
void linkBefore(E e, Node<E> succ) {
// assert succ != null;
final Node<E> pred = succ.prev;
final Node<E> newNode = new Node<>(pred, e, succ);
succ.prev = newNode;
if (pred == null)
first = newNode;
else
pred.next = newNode;
size++;
modCount++;
}
上面代碼中的node(int index)函數(shù)有一點(diǎn)小小的技巧,因?yàn)殒湵黼p向的,可以從開始往后找,也可以從結(jié)尾往前找,具體朝那個(gè)方向找取決于條件index < (size >> 1)
,也即是index是靠近前端還是后端。
/**
* 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;
}
}
從這里也可以看出,linkedList通過index檢索元素的效率沒有arrayList高
addAll()
addAll(index, c) 實(shí)現(xiàn)方式并不是直接調(diào)用add(index,e)來(lái)實(shí)現(xiàn),主要是因?yàn)樾实膯栴},另一個(gè)是fail-fast中modCount只會(huì)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()
為了讓GC更快可以回收放置的元素,需要將node之間的引用關(guān)系賦空。
/**
* 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++;
}
get相關(guān)的方法
【通過index獲取元素】
// Positional Access Operations
/**
* 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 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;
}
get(int index)
得到指定下標(biāo)處元素的引用,通過調(diào)用node(int index)方法實(shí)現(xiàn)
set()
將某個(gè)位置的元素重新賦值。
set(int index, E element)
方法將指定下標(biāo)處的元素修改成指定值,也是先通過node(int index)找到對(duì)應(yīng)下表元素的引用,然后修改Node中item的值
/**
* 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;
}
isElementIndex()
/**
* Tells if the argument is the index of an existing element.
*/
private boolean isElementIndex(int index) {
return index >= 0 && index < size;
}
isPositionIndex()
/**
* 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;
}
查找操作
查找操作的本質(zhì)是查找元素的下標(biāo)
indexOf
查找第一次出現(xiàn)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;
}
lastIndexOf
查找最后一次出現(xiàn)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;
}
本文摘自 :https://blog.51cto.com/u