目录
2.4.5 void add(int index, E element)
前言：
基于jdk1.8.
同上一篇ArrayList的源码。
两个半天撸完= =。DK加油
上一篇：https://blog.csdn.net/pmdream/article/details/106995088
1.LinkedList的复杂度？
get() 获取第几个元素，依次遍历，复杂度O(n)
add(E) 添加到末尾，复杂度O(1)
add(index, E) 添加第几个元素后，需要先查找到第几个元素，直接指针指向操作，复杂度O(n) （这个比较容易想错）
remove（）删除元素，直接指针指向操作，复杂度O(1)
2.源码分析
2.1 LinkedList的继承关系？
public class LinkedList<E>
extends AbstractSequentialList<E>
implements List<E>, Deque<E>, Cloneable, java.io.Serializable
{
//...
}对比ArrayList：
public class ArrayList<E> extends AbstractList<E>
implements List<E>, RandomAccess, Cloneable, java.io.Serializable
{
//...
}没有了快速访问，多了Deque接口。
因为实现了队列接口，所以可以让LinkedList作为双向队列。
和ArrayList不同的地方是，LinkedList继承的是AbstractSequentialList,AbstractSequentialList继承AbstractList，而ArrayList直接继承AbstractList。
2.2 属性
2.2.1 Node结构
因为不同于ArrayList，对于LinkedList每个元素都有着Node节点（这是一个内部类，有这前后指针）：
双向链表，那么顺序访问的效率很高，随机访问的效率低。因为通过索引访问，会比较索引值和链表长度的1/2，如果索引大，那么会从链表尾开始找，否则是头。所以才有了下面的first和last节点。
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;
}
} 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;2.3 构造函数
无参构造函数：
/**
* Constructs an empty list.
*/
public LinkedList() {
}有参构造函数：（因为链表形式不需要初始化空间给它）
/**
* 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);
}2.3.1 addAll(重磅)
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;
}
}
/**
* 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) {
//检查index是否合法
checkPositionIndex(index);
Object[] a = c.toArray();
int numNew = a.length;
if (numNew == 0)
return false;
//succ：待添加节点的位置。
//pred：待添加节点的前一个节点。
//新添加的元素的位置位于LinkedList最后一个元素的后面。
//如果index==size;说明此时需要添加LinkedList中的集合中的每一个元素都是在LinkedList
//最后面。所以把succ设置为空，pred指向尾节点。
//否则的话succ指向插入待插入位置的节点。pred指向succ节点的前一个节点。
Node<E> pred, succ;
if (index == size) {
//其实如果没有元素，也会进入第一个if判断
//说明想要插入的是在链表的末尾
succ = null;
pred = last;
} else {
//这种情况返回要插入的是哪个节点那里，但是肯定是插在succ之前，pred之后，所以接下来要操作pred
succ = node(index);
pred = succ.prev;
}
//循环a 每个节点存储着数组a的值，该节点的prev用来存储pred节点，next设置为空。接着判断一下该节点的前一个节点是否为
//空，如果为空的话，则把当前节点设置为头节点。否则的话就把当前节点的前一个节点的next值
//设置为当前节点。最后把pred指向当前节点，以便后续新节点的添加。
for (Object o : a) {
@SuppressWarnings("unchecked") E e = (E) o;
Node<E> newNode = new Node<>(pred, e, null);
if (pred == null)
//说明之前都是空的
first = newNode;
else
//那么如果pred不是空的了，把当前节点的前一个节点pred，的next指向新的内容
pred.next = newNode;
//pred指向newNode
pred = newNode;
}
if (succ == null) {
//新添加的节点位于LinkedList集合的最后一个元素的后面
//pred指向的是LinkedList中的最后一个元素
//所以把last指向pred
last = pred;
} else {
//中间位置插入的，相当于把一块内容插了进去，然后把前后指针都重新指向
pred.next = succ;
succ.prev = pred;
}
//加上新的集合大小
size += numNew;
//相当于整体链表修改的次数增加
modCount++;
return true;
}
//检查Pos是否合法，其实就是检查是否大于0并且在总大小内
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;
}
/**
* Returns the (non-null) Node at the specified element index.
* 最后返回一个
*/
Node<E> node(int index) {
// assert isElementIndex(index);
//与数组大小的一半进行比较
if (index < (size >> 1)) {
//如果index小于总大小的一半
//将first给x
Node<E> x = first;
//从头指针往后找
for (int i = 0; i < index; i++)
x = x.next;
return x;
} else {
Node<E> x = last;
//从尾指针开始往前找
for (int i = size - 1; i > index; i--)
x = x.prev;
return x;
}
}读了蛮久，总结就是有succ指针，指向当前index节点，pred是succ之前的节点，
那么想插入到index的时候，那么就得放到succ之前，pred之后，曹总pred指针呗~
最后再处理一下这两个的前后指针。
2.4 添加元素
2.4.1 add(E e)
/**
* 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.
当然连接到尾部了。
*/
void linkLast(E e) {
//这个last是内部属性，代表尾部节点指针
final Node<E> l = last;
//声明一个新的newNode，next指针指向null，前置指针指向之前的last
final Node<E> newNode = new Node<>(l, e, null);
last = newNode;
//这个书名之前的last是null，那么是相当于初始化，得吧first也得指向新的newNode
if (l == null)
first = newNode;
else
l.next = newNode;
size++;
modCount++;
}2.4.2 addAll(Collection<? extends E> c)
见上面构造函数。
2.4.3 addFirst(E e)
/**
* Inserts the specified element at the beginning of this list.
*
* @param e the element to add
*/
public void addFirst(E e) {
linkFirst(e);
}
/**
* 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++;
}基本思想一样，就是更改头部指针。
2.4.4 addLast(E 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);
}这个等同于add，只是没有返回添加称没成功的boolean
2.4.5 void add(int index, E element)
/**
* 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
//这种已经确保了，不会为null，因为如果链表长度只有1，那你却想插入5的index，就会报错的，运行期会报错
linkBefore(element, node(index));
}这里面需要注意linkBefore这个方法。node(index)找到了对应的要插入的那个节点的succ，当前index指向的节点。
/**
* 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;
//为什么会有判断pred == null？？接下来讲
if (pred == null)
first = newNode;
else
pred.next = newNode;
size++;
modCount++;
}在上面代码中，我不太明白为什么pred == null 有这个判断？
首先，进入这个方法的，肯定是有大小的。因为链表长度为1，但是比如我要插入5，就会报错。
public static void main(String[] args) {
//这样会报异常
LinkedList linkedList = new LinkedList();
linkedList.add("hh");
linkedList.add(5,"ss");
System.out.println(linkedList);
}那么怎么进入pred==null这个判断呢？
public static void main(String[] args) {
LinkedList linkedList = new LinkedList();
linkedList.add("hh");
linkedList.add(0,"ss");
System.out.println(linkedList);
}这种情况下，pred 就是空的。因为只有一个节点的时候，他的前后指针都是空的。
只不过我们会给first指针和last指针，但这不等于 node中的前后指针。
在进行这个操作之后，我们发现第一个节点有next，第二个节点有prev~
2.5 删除元素
2.5.1 remove()
/**
* 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;
}这个默认会删除掉头部的元素。
这边有一个点很可借鉴，当我们不用了一个元素的时候，需要把它的前后置镇 对于头结点就是next指针置为null，
这样可以帮助GC回收元素，因为就少了引用了。否则可能会持续的引用
2.5.2 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;
}如果是null直接用== 来判断相等，如果是对象类型使用equals。
unlink：
/**
* 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) {
//是头结点
first = next;
} else {
//否则，把之前节点的next指向next；并且释放x的前街店指针
prev.next = next;
x.prev = null;
}
if (next == null) {
//是尾结点
last = prev;
} else {
next.prev = prev;
x.next = null;
}
x.item = null;
size--;
modCount++;
return element;
}2.6 get
2.6.1 get(int index)
/**
* 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) {
//验证index合法性
checkElementIndex(index);
//使用node方法
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;
}
}2.6.2 getFirst()
/**
* 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;
}可以用属性中的first指针。
2.6.3 getLast()
/**
* 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;
}2.7 contains
/**
* Returns {@code true} if this list contains the specified element.
* More formally, returns {@code true} if and only if this list contains
* at least one element {@code e} such that
* <tt>(o==null ? e==null : o.equals(e))</tt>.
*
* @param o element whose presence in this list is to be tested
* @return {@code true} if this list contains the specified element
*/
public boolean contains(Object o) {
return indexOf(o) != -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;
}做了一遍循环。
2.8 set 设置对应index的节点的值
/**
* 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;
}返回oldVal
2.9 队列相关
2.9.1 peek
/**
* 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;
}返回头结点的值。（头结点为空，直接返回空）
2.9.2 poll
/**
* 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);
}移除头结点，并返回移除的头结点的值。
2.9.3 offer
/**
* 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);
}在链表的头部添加一个新的元素
2.9.4 push
/**
* 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);
}Push的话是添加头结点
2.9.5 pop
/**
* 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();
}2.10 序列化
2.10.1 writeObject
/**
* Saves the state of this {@code LinkedList} instance to a stream
* (that is, serializes it).
*
* @serialData The size of the list (the number of elements it
* contains) is emitted (int), followed by all of its
* elements (each an Object) in the proper order.
*/
private void writeObject(java.io.ObjectOutputStream s)
throws java.io.IOException {
// Write out any hidden serialization magic
s.defaultWriteObject();
// Write out size
s.writeInt(size);
// Write out all elements in the proper order.
for (Node<E> x = first; x != null; x = x.next)
s.writeObject(x.item);
}和ArrayList很像。
2.10.2 readObject
/**
* Reconstitutes this {@code LinkedList} instance from a stream
* (that is, deserializes it).
*/
@SuppressWarnings("unchecked")
private void readObject(java.io.ObjectInputStream s)
throws java.io.IOException, ClassNotFoundException {
// Read in any hidden serialization magic
s.defaultReadObject();
// Read in size
int size = s.readInt();
// Read in all elements in the proper order.
for (int i = 0; i < size; i++)
linkLast((E)s.readObject());
}
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