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JDK1.8 LinkedList源码分析

LinkedList简介

LinkedList 是一个继承于AbstractSequentialList的双向链表。它也可以被当作堆栈、队列或双端队列进行操作

LinkedList 实现 List 接口,能对它进行队列操作

LinkedList 实现 Deque 接口,即能将LinkedList当作双端队列使用

LinkedList 实现了Cloneable接口,即覆盖了函数clone(),能克隆

LinkedList 实现java.io.Serializable接口,这意味着LinkedList支持序列化,能通过序列化去传输

LinkedList 是非同步的

LinkedList属性

public class LinkedList<E>
    extends AbstractSequentialList<E>
    implements List<E>, Deque<E>, Cloneable, java.io.Serializable{   
    //当前有多少个节点
    transient int size = 0;
    //第一个节点
    transient Node<E> first;
    //最后一个节点
    transient Node<E> last;
}      

LinkedList构造方法

/**
     * 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);
    }
    /**
     * 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;
    }      

默认构造方法是空的,什么都没做,表示初始化的时候size为0,first和last的节点都为空:

另一个构造方法是带Collection值得对象作为入参的构造函数的,下面是执行逻辑:

  1. 使用this()调用默认的无参构造函数。
  2. 调用addAll()方法,传入当前的节点个数size,此时size为0,并将collection对象传递进去
  3. 检查index有没有数组越界的嫌疑
  4. 将collection转换成数组对象a
  5. 循环遍历a数组,然后将a数组里面的元素创建成拥有前后连接的节点,然后一个个按照顺序连起来。
  6. 修改当前的节点个数size的值
  7. 操作次数modCount自增1.

add方法

add(E e)方法

/**
     * 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++;
    }      

该方法直接将新增的元素放置链表的最后面,然后链表的长度(size)加1,修改的次数(modCount)加1

add(int index, E element)方法

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

指定位置往数组链表中添加元素

  1. 检查添加的位置index 有没有小于等于当前的长度链表size,并且要求大于0
  2. 如果是index是等于size,那么直接往链表的最后面添加元素,相当于调用add(E e)方法
  3. 如果index不等于size,则先是索引到处于index位置的元素,然后在index的位置前面添加新增的元素。

get方法

get方法

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

首先是判断索引位置有没有越界,确定完成之后开始遍历链表的元素,那么从头开始遍历还是从结尾开始遍历呢,这里其实是要索引的位置与当前链表长度的一半去做对比,如果索引位置小于当前链表长度的一半,否则从结尾开始遍历

getfirst方法

/**
     * 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;
    }      

直接将第一个元素返回

getlast方法

/**
     * 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;
    }      

直接将最后一个元素返回

remove方法

remove()方法

/**
     * 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();
    }      

remove方法本质调用的还是removeFirst方法(递归)

removeFirst()方法

/**
     * 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;
    }      

移除第一个节点,将第一个节点置空,让下一个节点变成第一个节点,链表长度减1,修改次数加1,返回移除的第一个节点。

removeLast()方法

/**
  * 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;
 }      

移除最后一个节点,将最后一个节点置空,最后一个节点的上一个节点变成last节点,链表长度减1,修改次数加1,返回移除的最后一个节点。

remove(int index)方法

/**
     * 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));
    }

    /**
     * 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;
    }      

先是检查移除的位置是否在链表长度的范围内,如果不在则抛出异常,根据索引index获取需要移除的节点,将移除的节点置空,让其上一个节点和下一个节点对接起来

set方法

/**
     * 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;
    }      

检查设置元素位然后置是否越界,如果没有,则索引到index位置的节点,将index位置的节点内容替换成新的内容element,同时返回旧值

clear方法

/**
     * 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++;
    }      

将所有链表元素置空,然后将链表长度修改成0,修改次数加1

push和pop方法

/**
     * 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();
    }      

push其实就是调用addFirst(e)方法,pop调用的就是removeFirst()方法

toArray方法

/**
     * 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 (Node<E> x = first; x != null; x = x.next)
            result[i++] = x.item;
        return result;
    }      

listIterator方法

/**
   * 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) {
      checkPositionIndex(index);
      return new ListItr(index);
  }