分析轮子- LinkedList.java

Posted 2021-01-05 godtrue

注：玩的是JDK1.7版本

一：先上类的继承结构图

二：再看一下他的底层实现数据结构

三：然后从源码中找点好玩的东西

 1）双向链表的结构构成元素，头指针、尾指针、节点信息（前向指针、后向指针、节点信息）

    /**
     * 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;

    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;
        }
    }

2）在链表尾部添加新节点，只需要改变链表尾指针的指针指向就可了，所以，性能相对 ArrayList.java 的数组拷贝会好很多

    /**
     * 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++;
    }

3）根据索引位置，获取节点，会遍历链表性能相对会差一些，并且我们也能看到遍历的时候也有一个优化，通过索引位置和链表长度的一半比较，决定从头遍历还是从尾遍历，这样能通过减少遍历长度加快遍历速度的作用，不过当索引位置在中间的时候，就不好玩了，无论从头或从尾遍历性能不会比较差

    /**
     * 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;
        }
    }

4）根据节点信息获取索引位置的方法，可以看到节点信息是允许为null的

    /**
     * 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;
    }

5）下面的三个方法，是根据索引位置，修改、删除索引位置的节点，都会遍历链表，所以，性能相对会差一些，ArrayListljava 是通过数组的索引位置直接操作的

    /**
     * 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;
    }

    /**
     * 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));
    }

    /**
     * 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));
    }

6）双向链表也可以当做双向队列使用