其他 源码解析 https://blog.csdn.net/qq_32726809/article/category/8035214
通过大体浏览源码，可知，Linkedlist的存储机构是一个链表
类的声明
public class LinkedList<E>
extends AbstractSequentialList<E>
implements List<E>, Deque<E>, Cloneable, java.io.Serializable
- AbstractSequentialList
- 顺序集合接口的最小实现
- 若是随机插入，则优先用AbstractList
- Deque
- 删除和插入的集合，双端对列
节点内部类
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;
}
}
next 指的是下一个节点，prev指的是上一个节点，item指的是当前节点的元素。可以将next和perv理解为指针。
属性
transient int size = 0;
transient Node<E> first;/*第一个节点*/
transient Node<E> last;/*最后一个节点*/
构造函数
public LinkedList(Collection<? extends E> c) {
this();
addAll(c);
}
按照集合的迭代器的元素获取构造函数
3方法
3.1linkFirst
把元素e作为第一个元素
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++;
}
将node节点上一个元素置为null，下一个元素为 原第一个元素，并将现在的元素赋予 first
3.2linkLast
把元素e多为最后一个元素
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++;
}
将当前元素新建一个node节点，把尾部的节点作为这个节点的上一个节点，把下一个节点置为null，把这个节点作为上一个节点的下一个节点，总的来说，就是把新节点与原来的节点连接起来.
3.3linkBefore
把元素e插入到 succ元素之前
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++;
}
类似于在两个火车车厢中间插入新的车厢，需要旧的车厢与新的车厢头尾相连，并改变车厢数
3.4unlinkFirst
将第一个元素移除链接
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;
}
3.5unlinkLast
将最后一个元素移除处链接
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;
}
3.6getFirst
获取第一个元素
public E getFirst() {
final Node<E> f = first;
if (f == null)
throw new NoSuchElementException();
return f.item;
}
3.7getLast
获取最后一个元素
public E getLast() {
final Node<E> l = last;
if (l == null)
throw new NoSuchElementException();
return l.item;
}
3.8removeFirst(),removeLast(),addFirst(E e),addLast(E e)
这些方法用的是上面已经讲过的方法
public E removeFirst() {
final Node<E> f = first;
if (f == null)
throw new NoSuchElementException();
return unlinkFirst(f);
}
/**
* 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);
}
/**
* Inserts the specified element at the beginning of this list.
*
* @param e the element to add
*/
public void addFirst(E e) {
linkFirst(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);
}
3.9contains
判断元素是否在集合中，其实就是判断节点的索引是否大于小于0
public boolean contains(Object o) {
return indexOf(o) != -1;
}
public int indexOf(Object o) {/*--------------1*/
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) {/*--------------2*/
if (o.equals(x.item))
return index;
index++;
}
}
return -1;
}
- 1处方法的是获取元素的所在位置
- 2处是遍历链条，对每个节点进行判断。
3.10add
public boolean add(E e) {
linkLast(e);
return true;
}
新添加的元素会被添加到最后
3.11 remove
通过遍历链表移除元素
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;
}
E unlink(Node<E> x) { /*-------------1*/
// 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处的函数是删除元素，然后将删除后的两端重新链接起来
3.12addAll
public boolean addAll(Collection<? extends E> c) {
return addAll(size, c);
}
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;
}
/*-----------------------1*/
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;
}
1处为核心代码，就是把集合转化的数组进行遍历，然后将元素新建成节点，再将节点链接起来
3.13clear
将所有连接元素设置为空
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++;
}