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LinkedList源码学习 + jdk11

程序员文章站 2022-04-19 18:23:35
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底层是使用双向链表实现的,插入删除容易,不需要扩容,不是线程安全的。
1.类定义

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

2.成员变量

	transient int size = 0;
    transient Node<E> first;//头结点
    transient Node<E> last;//尾节点 重写序列化算法

3.构造器

public LinkedList() {
}

public LinkedList(Collection<? extends E> c) {
    this();
    addAll(c);
}
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;
        }
    }

4.成员方法
add

public boolean add(E e) {
    linkLast(e);
    return true;
}

public void add(int index, E element) {
    checkPositionIndex(index);

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

public void addFirst(E e) {
	linkFirst(e);
}

public void addLast(E e) {
	linkLast(e);
}

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++;
}

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++;
}

remove

 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;
    }

get

public E get(int index) {
    checkElementIndex(index);
    return node(index).item;
}

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;
    }
}

参考