LinkedList源码分析
1.概述
2.源码分析
1.构造方法
//集合元素数量
transient int size = 0;
/**
* Pointer to first node.
* Invariant: (first == null && last == null) ||
* (first.prev == null && first.item != null)
*
* 意思是：集合为空或者
* 首元素的直接前驱是空并且首元素的值不为空
*
*/
//头结点
transient com.mrl.collection.linedlist.LinkedList.Node<E> first;
/**
* Pointer to last node.
* Invariant: (first == null && last == null) ||
* (last.next == null && last.item != null)
*/
//尾节点
transient com.mrl.collection.linedlist.LinkedList.Node<E> last;
/**
* Constructs an empty list.
*
* 空构造函数
*
*/
public LinkedList() {
}
/**
* Constructs a list containing the elements of the specified
* collection, in the order they are returned by the collection‘s
* iterator.
*
* 指定集合的构造
*
* @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();
//添加元素,插入链表
addAll(c);
}节点的结构
private static class Node<E> {
//元素
E item;
//直接后继
com.mrl.collection.linedlist.LinkedList.Node<E> next;
//直接前驱
com.mrl.collection.linedlist.LinkedList.Node<E> prev;
//构造
Node(com.mrl.collection.linedlist.LinkedList.Node<E> prev, E element, com.mrl.collection.linedlist.LinkedList.Node<E> next) {
this.item = element;
this.next = next;
this.prev = prev;
}
}2.新增
新增一个集合
/**
* 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.
*
* 以index为下标，插入集合c中所有元素
*
* @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) {
//检查是否越界 index>=0&&index<=size
checkPositionIndex(index);
//转化为数组
Object[] a = c.toArray();
//长度
int numNew = a.length;
if (numNew == 0)
return false;
// //index节点的前置节点，后置节点
com.mrl.collection.linedlist.LinkedList.Node<E> pred, succ;
//在链表末尾追加元素
if (index == size) {
//队尾节点的后置节点为null
succ = null;
//前置节点是队尾
pred = last;
} else {
//取出index节点，作为后置节点
succ = node(index);
//前置节点是，index节点的前一个节点
pred = succ.prev;
}
//for循环依次插入节点，而ArrayList是通过数组的拷贝
for (Object o : a) {
@SuppressWarnings("unchecked")
E e = (E) o;
//以前置节点，元素值，构建一个新节点
com.mrl.collection.linedlist.LinkedList.Node<E> newNode
= new com.mrl.collection.linedlist.LinkedList.Node<>(pred, e, null);
//前置节点是空，说明是头节点
if (pred == null)
//构造的新节点赋值给头节点
first = newNode;
else//否则前置节点的后置节点设置为新的节点
pred.next = newNode;
//步进，当前的节点设置为前置节点。为下次添加节点做准备
pred = newNode;
}
//循环结束后判断，如果后置节点是null，是在队尾追加的，
if (succ == null) {
last = pred;//设置尾节点
} else {
//是在队中插入的节点,
pred.next = succ; //更新前置节点的后置节点
succ.prev = pred;//更新后置节点的后置节点
}
/// 修改数量size
size += numNew;
//修改modCount
modCount++;
return true;
}
/**
* Returns the (non-null) Node at the specified element index.
*
* //根据index查出node
*/
com.mrl.collection.linedlist.LinkedList.Node<E> node(int index) {
// assert isElementIndex(index);
//折半判断index
if (index < (size >> 1)) {
com.mrl.collection.linedlist.LinkedList.Node<E> x = first;
for (int i = 0; i < index; i++)
x = x.next;
return x;
} else {
com.mrl.collection.linedlist.LinkedList.Node<E> x = last;
for (int i = size - 1; i > index; i--)
x = x.prev;
return x;
}
}小结：
1.链表批量增加是靠for循环依次增加的，ArrayList是靠System.arraycopy批量复制
2.通过下标获取某个node的时候，会判断index处于前半段还是后半段。提升查询效率
2.插入单个节点Node：
1.在尾部插入：
/**
* Appends the specified element to the end of this list.
* 尾部插入一个节点
* <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) {
//记录原尾部节点
final com.mrl.collection.linedlist.LinkedList.Node<E> l = last;
//以原来的尾部节点为前置节点构造新节点
final com.mrl.collection.linedlist.LinkedList.Node<E> newNode
= new com.mrl.collection.linedlist.LinkedList.Node<>(l, e, null);
//更新尾部节点
last = newNode;
if (l == null)
//如果尾部节点是null，即原来是空链表，额外更新头节点
first = newNode;
else
//更新原尾节点的后置节点为新节点
l.next = newNode;
//修改size
size++;
//修改modCount
modCount++;
}2.在指定下标插入：
/**
* 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).
* 在指定位置插入一个节点元素
* @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) {
//检查是否越界
checkPositionIndex(index);
//在末尾追加
if (index == size)
linkLast(element);
else
//中间插入
linkBefore(element, node(index));
}
/**
* Inserts element e before non-null Node succ.
*
* 在succ节点之前插入一个新节点
*
*/
void linkBefore(E e, com.mrl.collection.linedlist.LinkedList.Node<E> succ) {
// assert succ != null;
//保存后置节点的前置节点
final com.mrl.collection.linedlist.LinkedList.Node<E> pred = succ.prev;
//构建新节点
final com.mrl.collection.linedlist.LinkedList.Node<E> newNode = new com.mrl.collection.linedlist.LinkedList.Node<>(pred, e, succ);
//新节点是原succ节点的前置节点
succ.prev = newNode;
if (pred == null)//如果之前的前置节点是空,说明succ是原头结点。所以新节点是现在的头结点
first = newNode;
else//否则构建前置节点的后置节点为new
pred.next = newNode;
size++;
modCount++;
}3.删除
/**
* 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) {
//检查下标
checkElementIndex(index);
//删除节点
return unlink(node(index));
}
/**
* Unlinks non-null node x.
*/
E unlink(com.mrl.collection.linedlist.LinkedList.Node<E> x) {
// assert x != null;
//当前元素的节点值
final E element = x.item;
//当前元素的后置节点
final com.mrl.collection.linedlist.LinkedList.Node<E> next = x.next;
//当前元素的前置节点
final com.mrl.collection.linedlist.LinkedList.Node<E> prev = x.prev;
//当前元素的前置节点是null，当前节点是头节点
if (prev == null) {
//后置节点赋为头节点
first = next;
} else {
//否则，移动指针，前置节点的后置节点的指向后置节点
prev.next = next;
x.prev = null;//当前元素的前置节点删除
}
//后置节点为空,说明当前是尾节点
if (next == null) {
last = prev;
} else {
//否则移动指针
next.prev = prev;
x.next = null;
}
//清空当前元素值
x.item = null;
size--;
modCount++;
return element;
}删除指定元素
/**
* 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).
* 删除指定元素
* @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 (com.mrl.collection.linedlist.LinkedList.Node<E> x = first; x != null; x = x.next) {
if (x.item == null) {
unlink(x);
return true;
}
}
} else {
for (com.mrl.collection.linedlist.LinkedList.Node<E> x = first; x != null; x = x.next) {
if (o.equals(x.item)) {
unlink(x);
return true;
}
}
}
return false;
}删除会修改modCount,按照下标删，根据index先找到Node,然后去unlink这个node，按照元素删除，会先找链表是否有这个Node，找到去unlink。
4.修改
/**
* Replaces the element at the specified position in this list with the
* specified element.
* 修改指定位置上的元素
* @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);
com.mrl.collection.linedlist.LinkedList.Node<E> x = node(index);
E oldVal = x.item;
x.item = element;
return oldVal;
}5.查询
/**
* Returns the element at the specified position in this list.
* 得到元素
* @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;
}根据对象找到下标：
/**
* 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.
* 根据节点对象，查询下标
* @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 (com.mrl.collection.linedlist.LinkedList.Node<E> x = first; x != null; x = x.next) {
if (x.item == null)
return index;
index++;
}
} else {
for (com.mrl.collection.linedlist.LinkedList.Node<E> x = first; x != null; x = x.next) {
if (o.equals(x.item))
return index;
index++;
}
}
return -1;
}从尾部找：
/**
* 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.
* 从尾部查找
* @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 (com.mrl.collection.linedlist.LinkedList.Node<E> x = last; x != null; x = x.prev) {
index--;
if (x.item == null)
return index;
}
} else {
for (com.mrl.collection.linedlist.LinkedList.Node<E> x = last; x != null; x = x.prev) {
index--;
if (o.equals(x.item))
return index;
}
}
return -1;
}6.转为数组
/**
* 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 (com.mrl.collection.linedlist.LinkedList.Node<E> x = first; x != null; x = x.next)
result[i++] = x.item;
return result;
}7.遍历
for循环
for (int i = 0; i < linkedList2.size(); i++) {
System.out.println(linkedList2.get(i));
}foreach(迭代器实现)
for (Object o : linkedList2) {
System.out.println(o);
}迭代器：
for(Iterator iterator = linkedList2.iterator();iterator.hasNext();){
System.out.println(iterator.next());
}
/**
* 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) {
//从零开始
checkPositionIndex(index);
//返回迭代器,可以指定index，如果是iterator(),从0开始
return new com.mrl.collection.linedlist.LinkedList.ListItr(index);
}
private class ListItr implements ListIterator<E> {
//上一次返回的节点
private com.mrl.collection.linedlist.LinkedList.Node<E> lastReturned;
//要返回的下一个元素的节点
private com.mrl.collection.linedlist.LinkedList.Node<E> next;
//要返回的下一个元素的索引
private int nextIndex;
//modCount，并发访问异常
private int expectedModCount = modCount;
ListItr(int index) {
// assert isPositionIndex(index);
//索引==size，next为空，否则找到当前索引的节点
next = (index == size) ? null : node(index);
//下一个索引置为索引
nextIndex = index;
}
public boolean hasNext() {
return nextIndex < size;
}
public E next() {
//检查并发访问异常
checkForComodification();
if (!hasNext())
throw new NoSuchElementException();
//要返回的值赋值给lastReturned
lastReturned = next;
//移动指针，向后移动一位
next = next.next;
//索引++
nextIndex++;
//返回
return lastReturned.item;
}
public boolean hasPrevious() {
return nextIndex > 0;
}
public E previous() {
//检查
checkForComodification();
if (!hasPrevious())
throw new NoSuchElementException();
//next如果是null，则是尾节点。lastReturned赋值为尾节点
//否则lastReturned为next的prev节点
lastReturned = next = (next == null) ? last : next.prev;
nextIndex--;//索引--
return lastReturned.item;
}
public int nextIndex() {
return nextIndex;
}
public int previousIndex() {
return nextIndex - 1;
}
public void remove() {
checkForComodification();
if (lastReturned == null)
throw new IllegalStateException();
//上次返回的节点后面的节点做个备份
com.mrl.collection.linedlist.LinkedList.Node<E> lastNext = lastReturned.next;
//释放节点
unlink(lastReturned);
//这个条件不会走
if (next == lastReturned)
next = lastNext;
else
//索引--
nextIndex--;
//删除
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;
//尾部追加
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();
}
}总结：
LinekedList的遍历一定不能用for循环遍历，速度太慢，用foreach或者迭代器。
3.总结
原文：https://www.cnblogs.com/fantansticmrl/p/10300771.html