|
打算从这篇文章开始读jdk源码，先从比较简单的LinkedList开始吧。
1.继承关系?
先来看一下LinkedList的继承情况，可以看出来，LinkedList继承的是AbstractSequentialList
public class LinkedList<E>
extends AbstractSequentialList<E>
implements List<E>, Deque<E>, Cloneable, java.io.Serializable
{
...
}
之前面试被问到ArrayList和LinkedList的继承类有什么区别，那么现在就来看看ArrayList的继承情况吧。
public class ArrayList<E> extends AbstractList<E>
implements List<E>, RandomAccess, Cloneable, java.io.Serializable
{
...
}
从上面的信息可以看出来，ArrayList直接继承了AbstractList
那么可以总结如下：LinkedList继承的是AbstractSequentialList；ArrayList继承的是AbstractList
2.?属性?
接着，我们来看一下LinkedList的属性
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;
关于transient关键字，可以看看Java中transient关键字的详细总结_夏日清风-CSDN博客_transient关键字?????????
3.?构造函数?
接着来看构造函数，在代码的开头部分给了两个构造函数，分别是无参构造和有参构造
/**
* 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);
}
?无参构造函数什么都不做，有参构造函数的参数是集合类，其做法是首先调用无参构造，然后再将参数中的集合添加到链表中，使用的是addAll()方法
/**
* 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;
}
代码中的addAll()有两个函数签名，分别是
addAll(Collection<? extends E> c)和
addAll(int index, Collection<? extends E> c)
有惨构造函数调用的是第一个函数，然后第一个函数又调用了第二个函数，两个函数的返回值都是boolean。下面来具体看一下第二个函数做了什么。
该函数的第一个参数index就是集合中元素开始插入的位置，那么假如我们一开始size=0，然后你填写插入的第一个位置是1，那么就有问题了，所以需要先检查一下index是否越界，于是乎就首先调用了checkPositionIndex(int index)函数，该函数的具体内容如下
private void checkPositionIndex(int index) {
if (!isPositionIndex(index))
throw new IndexOutOfBoundsException(outOfBoundsMsg(index));
}
/**
* 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;
}
我们可以发现，这个函数的作用就是检查index是否大于等于0并且小于size，如果不在这个范围内，那么就抛出IndexOutOfBoundsExpection异常。
然后，将集合c转成Object数组，如果这个数组中没有元素，那么直接返回false。
接着就要判断，插入的位置是否是末尾的位置，假如(index==size)，那么就证明插入的位置是末尾的位置，否则就不是末尾的位置，我们需要找到插入的那个位置。代码中使用node(int index)来寻找这个位置，详细代码如下
/**
* 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;
}
}
这个函数的逻辑并不复杂，首先判断index是否小于size/2，假如是的话，那么从开头开始找比较近；假如不是的话，那么从结尾开始找比较近，然后返回相应位置的结点。
在addAll(int index, Collection<? extends E> c)函数中，使用succ指向index位置的结点，pred指向succ的前一个结点。
再接下来就是构建新链的过程了，其中有实例化很多新的结点，我们先来看一下Node类的定义
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;
}
}
pred==null对应的是空链表的情况，如果链表为空，那么就将first指向新的结点，否则就让pred.next=newNode，最后需要让pred=newNode，对应的代码如下
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;
}
最后，再将链表拼接上，如果succ==null，那么就代表原来是没有元素的；否则代表原来有元素，将新链和旧链链接即可。
if (succ == null) {
last = pred;
} else {
pred.next = succ;
succ.prev = pred;
}
最后修改链表的长度size和修改次数modCount。对于修改次数modCount的讲解可以看下面这篇文章
Java中modCount的作用？_Yanliang-CSDN博客_modcount
?4.?常用方法
在链表的头部插入一个结点。需要注意的是，在操作的过程中，如果first不为null的话，那么f和first指向的不是一个结点，first指向刚插入的那个结点，f指向插入后的第二个结点。
/**
* 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++;
}
在链表的尾部插入一个结点
/**
* 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++;
}
在succ前面插入一个结点
/**
* 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++;
}
删除头结点
/**
* 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;
}
删除尾结点
/**
* 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;
}
删除非空结点
/**
* 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) {
first = next;
} else {
prev.next = next;
x.prev = null;
}
if (next == null) {
last = prev;
} else {
next.prev = prev;
x.next = null;
}
x.item = null;
size--;
modCount++;
return element;
}
添加一个结点，可以发现，结点一般都是添加在末尾的
/**
* 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;
}
删除某个结点，如果我们给的参数是null，那么需要找到第一个值为null的结点，将其删除。删除成功返回true，否则返回false
/**
* 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;
}
删除所有结点
/**
* 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++;
}
在特定位置添加结点，首先要判断index是否合法，如果不合法的话就抛出IndexOutOfBoundsException异常
/**
* 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));
}
移除特定位置的结点，同样要先判断index是否合法
/**
* 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));
}
查找含有某个特定值的结点第一次出现的位置，从头开始遍历
/**
* 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;
}
查找含有某个特定值的结点最后一次出现的位置，从末尾开始遍历
/**
* 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;
}
注：并没有展示所有的方法，只是展示我认为比较重要的一些方法
5.?迭代器
/**
* Returns a list-iterator of the elements in this list (in proper
* sequence), starting at the specified position in the list.
* Obeys the general contract of {@code List.listIterator(int)}.<p>
*
* The list-iterator is <i>fail-fast</i>: if the list is structurally
* modified at any time after the Iterator is created, in any way except
* through the list-iterator's own {@code remove} or {@code add}
* methods, the list-iterator will throw a
* {@code ConcurrentModificationException}. Thus, in the face of
* concurrent modification, the iterator fails quickly and cleanly, rather
* than risking arbitrary, non-deterministic behavior at an undetermined
* time in the future.
*
* @param index index of the first element to be returned from the
* list-iterator (by a call to {@code next})
* @return a ListIterator of the elements in this list (in proper
* sequence), starting at the specified position in the list
* @throws IndexOutOfBoundsException {@inheritDoc}
* @see List#listIterator(int)
*/
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();
}
}
我们用的最多的就是hasNext()和next()这两个方法。使用迭代器遍历的一个例子如下
public class question2 {
public static void main(String[] args) {
LinkedList<Integer> list = new LinkedList<>();
list.add(1);
list.add(2);
list.add(3);
for(Iterator iter=list.listIterator(); iter.hasNext();){
System.out.println(iter.next());
}
}
}
6.?clone()和toArray()
clone()会返回一个克隆的LinkedList实例对象
/**
* Returns a shallow copy of this {@code LinkedList}. (The elements
* themselves are not cloned.)
*
* @return a shallow copy of this {@code LinkedList} instance
*/
public Object clone() {
LinkedList<E> clone = superClone();
// Put clone into "virgin" state
clone.first = clone.last = null;
clone.size = 0;
clone.modCount = 0;
// Initialize clone with our elements
for (Node<E> x = first; x != null; x = x.next)
clone.add(x.item);
return clone;
}
toArray()会返回Object[]类型的数组
/**
* Returns an array containing all of the elements in this list
* in proper sequence (from first to last element).
*
* <p>The returned array will be "safe" in that no references to it are
* maintained by this list. (In other words, this method must allocate
* a new array). The caller is thus free to modify the returned array.
*
* <p>This method acts as bridge between array-based and collection-based
* APIs.
*
* @return an array containing all of the elements in this list
* in proper sequence
*/
public Object[] toArray() {
Object[] result = new Object[size];
int i = 0;
for (Node<E> x = first; x != null; x = x.next)
result[i++] = x.item;
return result;
}
这是我第一次读源码，欢迎和我交流讨论
|