集合
LinkedList（有序的，线程不安全）
-
为什么是有序
数据结构是双向链表，但是first和last分别的头结点和未节点不会存放数据-及（first.pre==null;last.next==null）
-
为什么线程不安全
因为他的add和remove方法没有加锁，当多线程访问的时候会出现线程安全问题，可以通过 Collections.synchronizedList()解决，或者自己使用Vector，或者自己封装一下都行
-
内存结构
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; } }双向链表
linkedList_Pic1.png
- 添加数据
add(E e)
源码：
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++;
}
可以看到是直接添加到最后的。原理：创建一个节点（pre为改链表的last,next为null,element=e）,然后将last改为创建的节点newNode，last.next=newNode
add(int index, E element)
源码：
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;
}
}
/**
* 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++;
}
1，首先判断是不是添加到最后一个节点，如果添加到最后一各节点的话那么直接执行add(e)对应的底层调用linkLast（e）
2,不是最后一个节点，那么先获取这个index所在的节点（node(index)）
(1),直接折办（size>>1），在前面的顺序循环查找找到index,node.next,在后面倒叙查找index.pred
3,找到了这个index对应的节点，先获取他的pred(因为我们要插入到这个index对应的位置，让他本省往后移动)，然后创建新的节点，Node(pred=获取出来indexNode的pred,next就是indexNode,element就是我们传进来的element),然后将pred的next节点指向创建的新节点，这样就完成了添加到某个位置
remove(index)
源码：
/**
* 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;
}
1,直接找到first,然后取出他的next(作为替代新的first)，
2,先断老first的链(first.next=null)
3,清楚元素(first.item=null)，
4,然后将first的nextNode变成心得first(first=nextNode),
5,nextNode的pred是以前的first,因为LinkedList的first头部不存数据的，所以让nextNode.pred=null
removeLast
源码：
/**
* 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;
}
原理和remove（）方法差不多看看就知道了，脑海中一定要有内存模型的概念
remove(int index)
源码：
/**
* 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));
}
/**
* 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;
}
}
/**
* 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;
}
1 还是先去找index对应的节点
2，取出indexNode的predNode和nextNode
3，predNode的next指向nextNode,nextNode的pred指向predNode
