【源碼閱讀】Java集合之二 - LinkedList源碼深度解讀

Java 源碼閱讀的第一步是Collection框架源碼，這也是面試基礎中的基礎； 針對Collection的源碼閱讀寫一個系列的文章; 本文是第二篇LinkedList。 ---@pdai

JDK版本

JDK 1.8.0_110

概述總結

• LinkedList底層是經過雙向鏈表實現的，具體是first和last兩個node元素；
• 鏈表的特性決定了LinkedList按照index查找元素的效率沒有ArrayList高，可是add和remove操做效率會高不少；
• 根據index查找node, 由於鏈表雙向的，能夠從開始日後找，也能夠從結尾往前找，具體朝那個方向找取決於條件index < (size >> 1)，也便是index是靠近前端仍是後端;
• LinkedList同時實現了Deque接口，，也就是說它既能夠看做一個順序容器，又能夠看做一個隊列，同時又能夠看做一個棧。
• 當你須要使用棧或者隊列時，能夠考慮使用LinkedList，一方面是由於Java官方已經聲明不建議使用Stack類，更遺憾的是，Java里根本沒有一個叫作Queue的類（它是個接口名字）。
• 關於棧或隊列，如今的首選是ArrayDeque，它有着比LinkedList（看成棧或隊列使用時）有着更好的性能。
• LinkedList也採用了fail-fast的機制，經過記錄modCount參數來實現。在面對併發的修改時，迭代器很快就會徹底失敗，而不是冒着在未來某個不肯定時間發生任意不肯定行爲的風險;
• LinkedList的實現方式決定了全部跟下標相關的操做都是線性時間，而在首段或者末尾刪除元素只須要常數時間。爲追求效率LinkedList沒有實現同步（synchronized），若是須要多個線程併發訪問，能夠先採用Collections.synchronizedList()方法對其進行包裝。

類關係圖

LinkedList實現的接口和繼承的類以下：

public class LinkedList<E>
    extends AbstractSequentialList<E>
    implements List<E>, Deque<E>, Cloneable, java.io.Serializable
{
}

類的實現

底層數據結構

LinkedList底層經過雙向鏈表實現，雙向鏈表的每一個節點用內部類Node表示。LinkedList經過first和last引用分別指向鏈表的第一個和最後一個元素。注意這裏沒有所謂的啞元，當鏈表爲空的時候first和last都指向null。

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;

其中Node是私有的內部類：

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;
        }
    }

鏈表基礎數據結構

在解讀LinkedList源碼時，須要抓住兩點去思考，一個是底層的Node, 具體即是鏈表涉及的相關操做； 另外一個是LinkedList所實現接口，除了具有List特性以外，因爲其還實現了Dueue的接口,因此它還具有隊列的特性。從鏈表的角度，最重要的是鏈表的增刪操做， 以下

/**
 * Links e as first element.
 */
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++;
}

/**
 * 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++;
}

/**
 * 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++;
}

/**
 * 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;
}

/**
 * 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;
}

/**
 * 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;
}

總結下上面的方法：

• 上述方法分爲link和unlink兩大類, 從不一樣位置(頭，中間，尾部)添加映射爲不一樣的三個方法，這樣一共六個方法；
• 對首尾的操做（linkFirst，linkLast， unlinkFirst， unlinkLast），對LinkedList自己來講不會對暴露接口，在內部調用，因此是private的；
• link和unlink的操做本質是對Node的操做， 同時對size進行對應的加減；
• fail-fast機制一樣適用LinkedList, 因此對全部上述操做都會更新modCount；
• 對LinkedList的全部增長刪除操做本質是調用這些底層的對鏈表操做；

構造函數

LinkedList(Collection<? extends E> c)就是調用addAll()方法

/**
     * Constructs an empty list.
     */
    public LinkedList() {
    }

    /**
     * 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) {
        this();
        addAll(c);
    }

getFirst(), getLast()

獲取第一個元素， 和獲取最後一個元素：

/**
     * 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;
    }

removeFirest(), removeLast(), remove(e), remove(index)

remove()方法也有兩個版本，一個是刪除跟指定元素相等的第一個元素remove(Object o)，另外一個是刪除指定下標處的元素remove(int index)。

刪除元素 - 指的是刪除第一次出現的這個元素, 若是沒有這個元素，則返回false；判讀的依據是equals方法， 若是equals，則直接unlink這個node；因爲LinkedList可存放null元素，故也能夠刪除第一次出現null的元素；

/**
     * 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&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;
    }

remove(int index)使用的是下標計數， 只須要判斷該index是否有元素便可，若是有則直接unlink這個node。

/**
     * 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));
    }

刪除head元素：

/**
     * 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;
    }

刪除last元素：

/**
     * 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;
    }

add()

add()方法有兩個版本，一個是add(E e)，該方法在LinkedList的末尾插入元素，由於有last指向鏈表末尾，在末尾插入元素的花費是常數時間。只須要簡單修改幾個相關引用便可；另外一個是add(int index, E element)，該方法是在指定下表處插入元素，須要先經過線性查找找到具體位置，而後修改相關引用完成插入操做。

/**
     * 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;
    }
    
    /**
     * 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++;
    }

add(int index, E element), 當index==size時，等同於add(E e);
若是不是，則分兩步：1.先根據index找到要插入的位置,即node(index)方法；2.修改引用，完成插入操做。

/**
     * 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).
     *
     * @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) {
        checkPositionIndex(index);

        if (index == size)
            linkLast(element);
        else
            linkBefore(element, node(index));
    }

上面代碼中的node(int index)函數有一點小小的trick，由於鏈表雙向的，能夠從開始日後找，也能夠從結尾往前找，具體朝那個方向找取決於條件index < (size >> 1)，也便是index是靠近前端仍是後端。從這裏也能夠看出，linkedList經過index檢索元素的效率沒有arrayList高。

/**
     * 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;
        }
    }

addAll()

addAll(index, c) 實現方式並非直接調用add(index,e)來實現，主要是由於效率的問題，另外一個是fail-fast中modCount只會增長1次；

/**
     * 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.
     *
     * @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) {
        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;
        }

        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;
    }

clear()

爲了讓GC更快能夠回收放置的元素，須要將node之間的引用關係賦空。

/**
     * 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++;
    }

Positional Access Operations

經過index獲取元素

/**
     * Returns the element at the specified position in this list.
     *
     * @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;
    }

將某個位置的元素從新賦值：

/**
     * Replaces the element at the specified position in this list with the
     * specified element.
     *
     * @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);
        Node<E> x = node(index);
        E oldVal = x.item;
        x.item = element;
        return oldVal;
    }

將元素插入到指定index位置：

/**
     * 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).
     *
     * @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) {
        checkPositionIndex(index);

        if (index == size)
            linkLast(element);
        else
            linkBefore(element, node(index));
    }

刪除指定位置的元素：

/**
     * 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));
    }

其它位置的方法：

/**
     * Tells if the argument is the index of an existing element.
     */
    private boolean isElementIndex(int index) {
        return index >= 0 && index < size;
    }

    /**
     * Tells if the argument is the index of a valid position for an
     * iterator or an add operation.
     */
    private boolean isPositionIndex(int index) {
        return index >= 0 && index <= size;
    }

    /**
     * Constructs an IndexOutOfBoundsException detail message.
     * Of the many possible refactorings of the error handling code,
     * this "outlining" performs best with both server and client VMs.
     */
    private String outOfBoundsMsg(int index) {
        return "Index: "+index+", Size: "+size;
    }

    private void checkElementIndex(int index) {
        if (!isElementIndex(index))
            throw new IndexOutOfBoundsException(outOfBoundsMsg(index));
    }

    private void checkPositionIndex(int index) {
        if (!isPositionIndex(index))
            throw new IndexOutOfBoundsException(outOfBoundsMsg(index));
    }

Search Operations

查找操做的本質是查找元素的下標:

查找第一次出現的index, 若是找不到返回-1；

/**
     * 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) {
        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;
    }

查找最後一次出現的index, 若是找不到返回-1；

/**
     * 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&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 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 (Node<E> x = last; x != null; x = x.prev) {
                index--;
                if (x.item == null)
                    return index;
            }
        } else {
            for (Node<E> x = last; x != null; x = x.prev) {
                index--;
                if (o.equals(x.item))
                    return index;
            }
        }
        return -1;
    }

Queue operations

/**
     * Retrieves, but does not remove, the head (first element) of this list.
     *
     * @return the head of this list, or {@code null} if this list is empty
     * @since 1.5
     */
    public E peek() {
        final Node<E> f = first;
        return (f == null) ? null : f.item;
    }

    /**
     * Retrieves, but does not remove, 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 element() {
        return getFirst();
    }

    /**
     * Retrieves and removes the head (first element) of this list.
     *
     * @return the head of this list, or {@code null} if this list is empty
     * @since 1.5
     */
    public E poll() {
        final Node<E> f = first;
        return (f == null) ? null : unlinkFirst(f);
    }

    /**
     * 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();
    }

    /**
     * Adds the specified element as the tail (last element) of this list.
     *
     * @param e the element to add
     * @return {@code true} (as specified by {@link Queue#offer})
     * @since 1.5
     */
    public boolean offer(E e) {
        return add(e);
    }

Deque operations

/**
     * Inserts the specified element at the front of this list.
     *
     * @param e the element to insert
     * @return {@code true} (as specified by {@link Deque#offerFirst})
     * @since 1.6
     */
    public boolean offerFirst(E e) {
        addFirst(e);
        return true;
    }

    /**
     * Inserts the specified element at the end of this list.
     *
     * @param e the element to insert
     * @return {@code true} (as specified by {@link Deque#offerLast})
     * @since 1.6
     */
    public boolean offerLast(E e) {
        addLast(e);
        return true;
    }

    /**
     * Retrieves, but does not remove, the first element of this list,
     * or returns {@code null} if this list is empty.
     *
     * @return the first element of this list, or {@code null}
     *         if this list is empty
     * @since 1.6
     */
    public E peekFirst() {
        final Node<E> f = first;
        return (f == null) ? null : f.item;
     }

    /**
     * Retrieves, but does not remove, the last element of this list,
     * or returns {@code null} if this list is empty.
     *
     * @return the last element of this list, or {@code null}
     *         if this list is empty
     * @since 1.6
     */
    public E peekLast() {
        final Node<E> l = last;
        return (l == null) ? null : l.item;
    }

    /**
     * Retrieves and removes the first element of this list,
     * or returns {@code null} if this list is empty.
     *
     * @return the first element of this list, or {@code null} if
     *     this list is empty
     * @since 1.6
     */
    public E pollFirst() {
        final Node<E> f = first;
        return (f == null) ? null : unlinkFirst(f);
    }

    /**
     * Retrieves and removes the last element of this list,
     * or returns {@code null} if this list is empty.
     *
     * @return the last element of this list, or {@code null} if
     *     this list is empty
     * @since 1.6
     */
    public E pollLast() {
        final Node<E> l = last;
        return (l == null) ? null : unlinkLast(l);
    }

    /**
     * Pushes an element onto the stack represented by this list.  In other
     * words, inserts the element at the front of this list.
     *
     * <p>This method is equivalent to {@link #addFirst}.
     *
     * @param e the element to push
     * @since 1.6
     */
    public void push(E e) {
        addFirst(e);
    }

    /**
     * Pops an element from the stack represented by this list.  In other
     * words, removes and returns the first element of this list.
     *
     * <p>This method is equivalent to {@link #removeFirst()}.
     *
     * @return the element at the front of this list (which is the top
     *         of the stack represented by this list)
     * @throws NoSuchElementException if this list is empty
     * @since 1.6
     */
    public E pop() {
        return removeFirst();
    }

    /**
     * Removes the first occurrence of the specified element in this
     * list (when traversing the list from head to tail).  If the list
     * does not contain the element, it is unchanged.
     *
     * @param o element to be removed from this list, if present
     * @return {@code true} if the list contained the specified element
     * @since 1.6
     */
    public boolean removeFirstOccurrence(Object o) {
        return remove(o);
    }

    /**
     * Removes the last occurrence of the specified element in this
     * list (when traversing the list from head to tail).  If the list
     * does not contain the element, it is unchanged.
     *
     * @param o element to be removed from this list, if present
     * @return {@code true} if the list contained the specified element
     * @since 1.6
     */
    public boolean removeLastOccurrence(Object o) {
        if (o == null) {
            for (Node<E> x = last; x != null; x = x.prev) {
                if (x.item == null) {
                    unlink(x);
                    return true;
                }
            }
        } else {
            for (Node<E> x = last; x != null; x = x.prev) {
                if (o.equals(x.item)) {
                    unlink(x);
                    return true;
                }
            }
        }
        return false;
    }

參考文章

• Java LinkedList源碼剖析 結合源碼對LinkedList進行講解 http://www.cnblogs.com/CarpenterLee/p/5457150.html