Java中arraylist和linkedlist源代码分析与性能比較

Posted 2020-10-01 wzzkaifa

Java中arraylist和linkedlist源代码分析与性能比較

1，简单介绍

在java开发中比較经常使用的数据结构是arraylist和linkedlist，本文主要从源代码角度分析arraylist和linkedlist的性能。

2，arraylist源代码分析

         Arraylist底层的数据结构是一个对象数组。有一个size的成员变量标记数组中元素的个数，例如以下图：

     * The array buffer into which the elements of the ArrayList are stored.
     * The capacity of the ArrayList is the length of this array buffer.
     */
    private transient Object[] elementData;

    /**
     * The size of the ArrayList (the number of elements it contains).
     *
     * @serial
     */
    private int size;

        在构造函数中Arraylist初始化为一个长度为10的对象数组。例如以下：    

    /**
     * Constructs an empty list with the specified initial capacity.
     *
     * @param  initialCapacity  the initial capacity of the list
     * @throws IllegalArgumentException if the specified initial capacity
     *         is negative
     */

    public ArrayList(int initialCapacity) {
        super();
        if (initialCapacity < 0)
            throw new IllegalArgumentException("Illegal Capacity: "+
                                               initialCapacity);
        this.elementData = new Object[initialCapacity];
    }

    /**
     * Constructs an empty list with an initial capacity of ten.
     */
    public ArrayList() {
        this(10);
    }

        Arraylist在添加数据时，首先推断数组是否超过原始分配数组的长度。假设超过。则通过数组复制的形式扩大数组然后再添加数组元素。时间复杂度处于O(1)到O(n)之间。源代码例如以下：

    /**
     * Appends the specified element to the end of this list.
     *
     * @param e element to be appended to this list
     * @return <tt>true</tt> (as specified by {@link Collection#add})
     */
    public boolean add(E e) {
        ensureCapacityInternal(size + 1);  // Increments modCount!!
        elementData[size++] = e;
        return true;
    }

    private void ensureCapacityInternal(int minCapacity) {
        modCount++;
        // overflow-conscious code
        if (minCapacity - elementData.length > 0)
            grow(minCapacity);
    }

   /**
     * Increases the capacity to ensure that it can hold at least the
     * number of elements specified by the minimum capacity argument.
     *
     * @param minCapacity the desired minimum capacity
     */
    private void grow(int minCapacity) {
        // overflow-conscious code
        int oldCapacity = elementData.length;
        int newCapacity = oldCapacity + (oldCapacity >> 1);
        if (newCapacity - minCapacity < 0)
            newCapacity = minCapacity;
        if (newCapacity - MAX_ARRAY_SIZE > 0)
            newCapacity = hugeCapacity(minCapacity);
        // minCapacity is usually close to size, so this is a win:
        elementData = Arrays.copyOf(elementData, newCapacity);
    }

     Arraylist在删除数据时，首先会推断数组是否越界，然后会做一个数组从后向前复制的操作，时间复杂度是O(N)，源代码例如以下图：

    /**
     * Removes the element at the specified position in this list.
     * Shifts any subsequent elements to the left (subtracts one from their
     * indices).
     *
     * @param index the index of the element to be removed
     * @return the element that was removed from the list
     * @throws IndexOutOfBoundsException {@inheritDoc}
     */
    public E remove(int index) {
        rangeCheck(index);

        modCount++;
        E oldValue = elementData(index);

        int numMoved = size - index - 1;
        if (numMoved > 0)
            System.arraycopy(elementData, index+1, elementData, index,
                             numMoved);
        elementData[--size] = null; // Let gc do its work

        return oldValue;
    }

Arraylist在改动数据时，首先推断第i个元素是否越界，然后直接做赋值操作。时间复杂度是O(1)。

    /**
     * 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) {
        rangeCheck(index);

        E oldValue = elementData(index);
        elementData[index] = element;
        return oldValue;
    }

Arraylist在获取数据时，首先判读第i个元素是否越界，然后获取对象数组中的第i个元素时间复杂度是O(1)。源代码例如以下图：

    /**
     * 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) {
        rangeCheck(index);

        return elementData(index);
    }

3，linkedlist源代码分析

                Linkedlist的用到的底层数据结构是双向链表，数据结构例如以下图：

    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;
        }
    }

Linkedlist的成员变量主要有三个，size表示链表的长度，first指向链表的头部，last指向链表的尾部，源代码例如以下图：

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

Linkedlist的添加操作。仅仅是在链表的尾部添加一个节点，时间复杂度是O(1)。源代码例如以下图：

    /**
     * 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++;
    }

Linkedlist的删除操作。首先推断删除的位置是否越界。然后找到第i个元素，最后删除第一个元素，由于在删除的时候要依据元素的位置获取元素，所以时间复杂度是O(N)。源代码例如以下：

    /**
     * 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));
    }

</pre><pre code_snippet_id="1635127" snippet_file_name="blog_20160405_24_2093084" name="code" class="java">    /**
     * 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;
        }
    }

    /**
     * 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;
    }

        Linkedlist的改动操作。首先是推断数组是否越界。然后获取当前位置的元素。最后做改动操作，因为在获取当前位置的元素时，须要遍历链表。所以时间复杂度是O(N)，源代码例如以下：

 

    /**

     * 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;
    }

    /**
     * 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;
        }
    }

        Linkedlist的获取元素操作。首先是推断数组是否越界，然后获取当前位置的元素，因为在获取当前位置的元素时，须要遍历链表，所以时间复杂度是O(N)，源代码例如以下：

    /**
     * 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。arraylist和linkedlist的效率分析

	Add()	Remove(int i)	Set(int i, E e)	Get(int i)
Arraylist	O(1)-O(N)	O(N)	O(1)	O(1)
Linkedlist	O(1)	O(N)	O(N)	O(N)