本文為部落格園作者所寫： 一寸HUI，個人部落格地址：https://www.cnblogs.com/zsql/
簡單的一個類就直接說了。LinkedList 的底層結構是一個帶頭/尾指標的雙向連結串列，可以快速的對頭/尾節點 進行操作，它允許插 入所有元素，包括 null。 相比陣列（這裡可以對比ArrayList原始碼分析進行檢視），連結串列的特點就是在指定位置插入和刪除元素的效率較高，但是查詢的 效率就不如陣列那麼高了。如果熟悉雙向連結串列這個資料結構，其實就很簡單了，無非就是實現一些資料的新增，刪除，查詢，遍歷等功能，雙向連結串列的結構圖如下：
每一個資料（節點）都包含3個部分，一個是資料本身item，一個是指向下一個節點的next指標，還有就是指向上一個節點的prev指標，另外，雙向連結串列還有一個 first 指標，指向頭節點，和 last 指標，指向尾節點。，在LinkedList類中通過私有的靜態內部類Node作為每一個資料的封裝。具體實現如下：
private static class Node<E> { //這個類就是用來封裝雙向連結串列中的每一個資料，也是上圖中的每一個框
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;
}
}
接下看看LinkList類的定義：
public class LinkedList<E>
extends AbstractSequentialList<E> //繼承的類
implements List<E>, Deque<E>, Cloneable, java.io.Serializable //實現的各種介面
{}
接下來看看LinkedList這個類的一些屬性：就三個屬性，一個用來記錄雙向連結串列的大小，一個是first節點用來指向連結串列的頭，last用來指向連結串列的尾
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;
在看看構造方法：
/**
* Constructs an empty list.
*/
public LinkedList() { //空參構造
}
/**
* 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) { //通過已有的集合進行構造
this();
addAll(c); //使用addAll（）方法把集合中的資料生產LinkedList
}
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(); //把集合轉為陣列
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) { //對陣列進行遍歷，對每一個元素都封裝成Node並新增到LinkedList中
@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;
}
接下來看看LinkedList的基本操作，新增，刪除，遍歷，查詢等
先看新增，從雙向連結串列的結構來看，新增元素可以在連結串列的頭、尾、以及中間的任意位置新增新的元素。因為 LinkedList 有頭指標和尾指標，所以在表頭或表尾進 行插入元素只需要 O(1) 的時間，而在指定位置插入元素則需要先遍歷一下連結串列， 所以複雜度為 O(n)。首先看看在頭部新增元素：
看圖可以看出，只要把first指向新的node，新的node的next指向原先firt指向的node，再把原先first指向的node的prev指向新的node就可以了。
/**
* Links e as first element.
*/
private void linkFirst(E e) {
final Node<E> f = first; //使用臨時node
final Node<E> newNode = new Node<>(null, e, f); //封裝新的node，並把新node的nex指向f
first = newNode;
if (f == null) //判斷first是否為空
last = newNode;
else
f.prev = newNode; //把f的prev指向新的node
size++; //連結串列長度加1
modCount++; //記錄連結串列被修改的次數
}
在看看在尾部新增，其實和在頭部新增一樣，只是把first換成了last，邏輯一樣
/**
* 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++;
}
再看看在中間的任意位置新增：
這個相對來說複雜點點，修改新增前後node的next和prev的指向，修改的相對來說多點點
/**
* Inserts element e before non-null Node succ.
*/
void linkBefore(E e, Node<E> succ) { //表示在在succ節點前面新增e元素
// assert succ != null;
final Node<E> pred = succ.prev; //獲取succ的前面節點
final Node<E> newNode = new Node<>(pred, e, succ); //把e封裝成節點，並把prev指向succ前面節點，把next指向succ節點
succ.prev = newNode; //然後把succ的prev指向新的節點
if (pred == null)
first = newNode;
else
pred.next = newNode; //把succ的前節點的next只想新的節點
size++; //連結串列長度+1
modCount++; //修改次數+1
}
新增說完了，就說說刪除，其實也很簡單
刪除也是分為從頭部、尾部、中間位置刪除
先看看從first位置刪除
/**
* Unlinks non-null first node f.
*/
private E unlinkFirst(Node<E> f) {
// assert f == first && f != null;
final E element = f.item; //獲取first中間的元素，用於後面的返回
final Node<E> next = f.next; //獲取f的next節點
f.item = null;
f.next = null; // help GC 清除
first = next; //把first指向f的next
if (next == null)
last = null;
else
next.prev = null; //清除
size--; //連結串列長度-1
modCount++; //修改次數+1
return element;
}
看了從頭部刪除，其實尾部刪除也差不多
/**
* 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;
}
在看看從指定位置刪除吧
/**
* Unlinks non-null node x.
*/
E unlink(Node<E> x) {
// assert x != null;
final E element = x.item; //獲取該節點的值
final Node<E> next = x.next; //獲取該節點的next節點
final Node<E> prev = x.prev; //獲取該節點的prev節點
if (prev == null) { //把該節點的前節點的next指向該節點的next節點，並清除該節點的prev指向
first = next;
} else {
prev.next = next;
x.prev = null;
}
if (next == null) { //把該節點的next節點的prev指向該節點的prev節點，並清除該節點的next指向
last = prev;
} else {
next.prev = prev;
x.next = null;
}
x.item = null; //清除
size--; //連結串列長度-1
modCount++; //修改次數+1
return element;
}
看完增刪，那就繼續看查相關的方法，也有從頭，尾相關的查詢方法，都很簡單，做判斷，然後查詢
/**
* 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;
}
/**
* 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;
}
當然還有指定index查詢的
/**
* Returns the (non-null) Node at the specified element index.
*/
Node<E> node(int index) {
// assert isElementIndex(index);
//判斷index是在連結串列的前半段還是在後半段，如果在前半段就從first向後遍歷，否則使用last向前遍歷
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;
}
}
其實基本知道了上面的方法基本對雙向連結串列有了一定的熟悉，當然LinkedList還有很多其他的方法，不過很多都是基於上面這些方法的一些封裝，例如：
/**
* 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);
}
/**
* 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);
}
/**
* 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);
}
/**
* 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;
}
/**
* 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;
}
/**
* 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++;
}
/**
* 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 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) { //查詢元素o是否在連結串列中，並返回index，沒找到返回-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) {
if (o.equals(x.item))
return index;
index++;
}
}
return -1;
}
到這裡本文就結束了了，如果想知道LinkedList的更多方法，建議去看原始碼