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LinkedList源码分析

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LinkedList源码分析

LinkedList底层数据结构是双向链表,它同时实现了ListDeque两个接口,插入和删除元素的时间复杂度均为O(1), 相比于ArrayList它在插入和删除元素操作上具有明显优势,同样它也是非线程安全的容器。由于实现了deque接口,因此linkedList也可以用于实现队列

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成员变量

transient int size = 0; //容器中元素数量

transient Node<E> first; //指向第一个元素的指针

transient Node<E> last;//指向最后一个元素的指针

Node类

private static class Node<E> {
        E item; //链表节点
        Node<E> next; //后驱指针
        Node<E> prev; //前继指针

        Node(Node<E> prev, E element, Node<E> next) {
            this.item = element;
            this.next = next;
            this.prev = prev;
        }
    }

添加元素

//在链表头部插入元素
private void linkFirst(E e) {
        final Node<E> f = first;
        final Node<E> newNode = new Node<>(null, e, f);
        first = newNode;
        //链表只有一个元素
        if (f == null)
            last = newNode;
        else
           // newNode成为第一个节点
            f.prev = newNode;
        size++;
        modCount++;
    }

//获取指定位置的元素
Node<E> node(int index) {
        // assert isElementIndex(index);
        //比较index位置与链表中间节点,确定左边区间还是右区间,减少遍历
        if (index < (size >> 1)) {
            Node<E> x = first;
            for (int i = 0; i < index; i++)
                x = x.next;
            return x;
        } else {
           //在右区间查找
            Node<E> x = last;
            for (int i = size - 1; i > index; i--)
                x = x.prev;
            return x;
        }
    }

//确定元素在链表中的位置
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 {
            //从首节点开始逐一遍历
            for (Node<E> x = first; x != null; x = x.next) {
                if (o.equals(x.item))
                    return index;
                index++;
            }
        }
        return -1;
    }

//在链表表尾添加元素
public boolean add(E e) {
        linkLast(e);
        return true;
    }

void linkLast(E e) {
        final Node<E> l = last;
        //新增节点
        final Node<E> newNode = new Node<>(l, e, null);
        last = newNode;
        if (l == null)
            first = newNode;
        else
            //新节点成为尾节点
            l.next = newNode;
        size++;
        modCount++;
    }

//在节点前添加元素
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);
        //修改指针指向
        succ.prev = newNode;
        if (pred == null)
            first = newNode;
        else
            pred.next = newNode;
        size++;
        modCount++;
    }

public void add(E e) {
            checkForComodification();
            lastReturned = null;
            //当前只有一个元素
            if (next == null)
                //相当于在表尾添加元素
                linkLast(e);
            else
                linkBefore(e, next);
            nextIndex++;
            expectedModCount++;
        }

删除节点

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 {
            //从第一个节点开始遍历
            for (Node<E> x = first; x != null; x = x.next) {
                if (o.equals(x.item)) {
                    //修改指针指向
                    unlink(x);
                    return true;
                }
            }
        }
        return false;
    }

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;
    }

标签:分析,Node,prev,LinkedList,++,next,源码,newNode,null
来源: https://www.cnblogs.com/crstly/p/15733697.html