LinkedList源码分析

来源：互联网发布：windows thin 编辑：程序博客网时间：2024/05/22 01:59

Java版本：

java version "1.7.0_79"Java(TM) SE Runtime Environment (build 1.7.0_79-b15)Java HotSpot(TM) 64-Bit Server VM (build 24.79-b02, mixed mode)

LinkedList由双向链表实现。
LinkedList主要成员变量：

    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;    //核心数据结构代码：    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 ;        }    }

通过定义的内部类，包括三个成员变量，数据对象，前一个节点对象，以及后一个节点对象。

add()/get()/remove()方法举例：

    /**     * 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.     * 首先获得最后一个元素，新建一个节点newNode，     * 如果最后一个元素为空，则将newNode赋值给first     * 否则将newNode赋值给最后一个元素的next     */    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++;    }    /**     * 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}     * 检查index合法性，调用node方法     */    public E get( int index ) {        checkElementIndex(index);        return node(index).item ;    }    /**     * Returns the (non -null) Node at the specified element index.     * 采用二分，如果index小于size >> 1(size/2),那么将从第一个元素开始往前推     * 否则，从最后一个元素往后推     */    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 ;        }    }    /**     * 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}     * 检查index合法性，根据index获得对应的Node元素作为参数，调用unlink方法     */    public E remove(int index) {        checkElementIndex(index);        return unlink(node(index));    }    /**     * Unlinks non -null node x.     * 获得Node对象的element，前一个节点prev及后一个节点next     * 处理x的prev节点：     * 如果移除的元素为首元素(即prev为null)，那么直接将节点next赋给first     * 否则，将next赋值给prev.next，并将x.prev置空     * 处理x的next节点：     * 如果移除的元素为尾元素(即next为null)，那么直接将节点prev赋值给last     * 否则，将prev赋值给next.prev，并将x.next置空     */    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 ;    }

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