LinkedList 的實現原理

本文爲博客園做者所寫： 一寸HUI，我的博客地址：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/尾指針的雙向鏈表，能夠快速的對頭/尾節點 進行操做，它容許插 入全部元素，包括 null。 相比數組（這裏能夠對比ArrayList源碼分析進行查看），鏈表的特色就是在指定位置插入和刪除元素的效率較高，可是查找的 效率就不如數組那麼高了。若是熟悉雙向鏈表這個數據結構，其實就很簡單了，無非就是實現一些數據的添加，刪除，查詢，遍歷等功能，雙向鏈表的結構圖以下：

 

 

每個數據（節點）都包含3個部分，一個是數據自己item，一個是指向下一個節點的next指針，還有就是指向上一個節點的prev指針，另外，雙向鏈表還有一個 first 指針，指向頭節點，和 last 指針，指向尾節點。，在LinkedList類中經過私有的靜態內部類Node做爲每個數據的封裝。具體實現以下：

private static class Node<E> { //這個類就是用來封裝雙向鏈表中的每個數據，也是上圖中的每個框 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 //實現的各類接口 {}

 

 

 接下來看看LinkedList這個類的一些屬性：就三個屬性，一個用來記錄雙向鏈表的大小，一個是first節點用來指向鏈表的頭，last用來指向鏈表的尾

   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) { //經過已有的集合進行構造 this(); addAll(c); //使用addAll（）方法把集合中的數據生產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(); //把集合轉爲數組 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) { //對數組進行遍歷，對每個元素都封裝成Node並添加到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的基本操做，添加，刪除，遍歷，查詢等

先看添加，從雙向鏈表的結構來看，添加元素能夠在鏈表的頭、尾、以及中間的任意位置添加新的元素。由於 LinkedList 有頭指針和尾指針，因此在表頭或表尾進 行插入元素只須要 O(1) 的時間，而在指定位置插入元素則須要先遍歷一下鏈表， 因此複雜度爲 O(n)。首先看看在頭部添加元素：

 

 

 看圖能夠看出，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就能夠了。

/** * 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是否爲空 last = newNode; else f.prev = newNode; //把f的prev指向新的node size++; //鏈表長度加1 modCount++; //記錄鏈表被修改的次數 }

在看看在尾部添加，其實和在頭部添加同樣，只是把first換成了last，邏輯同樣

/** * 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添加：

 

 

 這個相對來講複雜點點，修改添加先後node的next和prev的指向，修改的相對來講多點點

/** * Inserts element e before non-null Node succ. */
    void linkBefore(E e, Node<E> succ) { //表示在在succ節點前面添加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; //而後把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++; //鏈表長度+1 modCount++; //修改次數+1 }

添加說完了，就說說刪除，其實也很簡單

 

 

 刪除也是分爲從頭部、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--; //鏈表長度-1 modCount++; //修改次數+1 return element; }

看了從頭部刪除，其實尾部刪除也差很少

 /** * 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--; //鏈表長度-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，作判斷，而後查詢

/** * 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; }

固然還有指定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是在鏈表的前半段仍是在後半段，若是在前半段就從first向後遍歷，不然使用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鏈表有了必定的熟悉，固然LinkedList還有不少其餘的方法，不過不少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&nbsp;?&nbsp;get(i)==null&nbsp;:&nbsp;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&nbsp;?&nbsp;get(i)==null&nbsp;:&nbsp;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) { //查找元素o是否在鏈表中，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; }

到這裏本文就結束了了，若是想知道LinkedList的更多方法，建議去看源碼