LinkedList源码解析

Posted 2021-02-20 zerodslearnjava

LinkedList

1 类图

2 字段和内部类

    // 链表的大小
    transient int size = 0;
    // 指向第一个节点
    transient Node<E> first;
    // 指向最后一个节点
    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;
        }
    }

3 构造器

    public LinkedList() {
    }
    
    public LinkedList(Collection<? extends E> c) {
        this();
        addAll(c);
    }
    
    public boolean addAll(Collection<? extends E> c) {
        return addAll(size, c);
    }
    
    public boolean addAll(int index, Collection<? extends E> c) {
        checkPositionIndex(index);

        Object[] a = c.toArray();
        int numNew = a.length;
        if (numNew == 0)
            return false;

        Node<E> pred, succ;
        if (index == size) {
            succ = null;
            pred = last;
        } else {
            succ = node(index);
            pred = succ.prev;
        }

        for (Object o : a) {
            @SuppressWarnings("unchecked") E e = (E) o;
            Node<E> newNode = new Node<>(pred, e, null);
            if (pred == null)
                first = newNode;
            else
                pred.next = newNode;
            pred = newNode;
        }

        if (succ == null) {
            last = pred;
        } else {
            pred.next = succ;
            succ.prev = pred;
        }

        size += numNew;
        modCount++;
        return true;
    }
    
    private void checkPositionIndex(int index) {
        if (!isPositionIndex(index))
            throw new IndexOutOfBoundsException(outOfBoundsMsg(index));
    }
    
    private boolean isPositionIndex(int index) {
        return index >= 0 && index <= size;
    }

4 增

4.1 addFirst

    /**
     * 添加元素到头部
     */
    public void addFirst(E e) {
        linkFirst(e);
    }
    
    private void linkFirst(E e) {
        // 获取当前头部元素
        final Node<E> f = first;
        // 构造节点，前置节点为空，后置节点为当前头节点
        final Node<E> newNode = new Node<>(null, e, f);
        // 设置头节点的引用为新的节点
        first = newNode;
        // 如果当前为空链表，那么尾节点=头节点
        if (f == null)
            last = newNode;
        // 如果不是空列表，那么旧的头部节点的前置节点设置为新的头节点
        else
            f.prev = newNode;
        size++;
        modCount++;
    }

4.2 addLast

    // 添加元素到链表尾部
    public void addLast(E e) {
        linkLast(e);
    }
    
    public boolean add(E e) {
        linkLast(e);
        return true;
    }
    
    void linkLast(E e) {
        // 先获取尾节点
        final Node<E> l = last;
        // 创建尾部节点
        final Node<E> newNode = new Node<>(l, e, null);
        // 修改尾部节点引用
        last = newNode;
        // 如果是空链表，则头节点=尾节点
        if (l == null)
            first = newNode;
        // 非空列表，则修改尾节点的后置节点引用
        else
            l.next = newNode;
        size++;
        modCount++;
    }

4.3 add(int index, E element)

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

        if (index == size)
            linkLast(element);
        else
            linkBefore(element, node(index));
    }
    
    /**
     * Inserts element e before non-null Node succ.
     */
    void linkBefore(E e, Node<E> succ) {
        // 找到插入位置节点的前置节点prev
        // prev->new Node(e)->succ
        final Node<E> pred = succ.prev;
        final Node<E> newNode = new Node<>(pred, e, succ);
        succ.prev = newNode;
        if (pred == null)
            first = newNode;
        else
            pred.next = newNode;
        size++;
        modCount++;
    }
    
    /**
     * Returns the (non-null) Node at the specified element index.
     */
    Node<E> node(int index) {
        // assert isElementIndex(index);
        
        // 如果是在前半部分，从头节点往尾节点遍历
        // 否则，从尾节点向头节点遍历
        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;
        }
    }

5 删

    public E remove(int index) {
        checkElementIndex(index);
        return unlink(node(index));
    }
    
    E unlink(Node<E> x) {
        final E element = x.item;
        // 找到对应索引的节点的前置和后置节点
        final Node<E> next = x.next;
        final Node<E> prev = x.prev;

        // 如果前置节点为空，那么头节点=后置节点
        // 否则前置节点的next变成后置节点,需要删除节点的前置节点变成空
        // prev->delete->next
        // prev->next
        if (prev == null) {
            first = next;
        } else {
            prev.next = next;
            x.prev = null;
        }

        // 如果后置节点为空，那么尾节点=前置节点
        // 否则后置节点的prev变成前置节点,需要删除节点的后置节点变成空
        // prev->delete->next
        // prev->next
        if (next == null) {
            last = prev;
        } else {
            next.prev = prev;
            x.next = null;
        }

        // 指向的引用置为空
        x.item = null;
        size--;
        modCount++;
        return element;
    }

6 改

    public E set(int index, E element) {
        checkElementIndex(index);
        Node<E> x = node(index);
        E oldVal = x.item;
        x.item = element;
        return oldVal;
    }

7 查

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

8 遍历和迭代器

避免每次调用get(i),这样的性能开销较高

    xuexprivate class ListItr implements ListIterator<E> {
        private Node<E> lastReturned;
        private Node<E> next;
        private int nextIndex;
        private int expectedModCount = modCount;

        ListItr(int index) {
            // assert isPositionIndex(index);
            next = (index == size) ? null : node(index);
            nextIndex = index;
        }

        public boolean hasNext() {
            return nextIndex < size;
        }

        public E next() {
            checkForComodification();
            if (!hasNext())
                throw new NoSuchElementException();

            lastReturned = next;
            next = next.next;
            nextIndex++;
            return lastReturned.item;
        }

        public boolean hasPrevious() {
            return nextIndex > 0;
        }

        public E previous() {
            checkForComodification();
            if (!hasPrevious())
                throw new NoSuchElementException();

            lastReturned = next = (next == null) ? last : next.prev;
            nextIndex--;
            return lastReturned.item;
        }

        public int nextIndex() {
            return nextIndex;
        }

        public int previousIndex() {
            return nextIndex - 1;
        }

        public void remove() {
            checkForComodification();
            if (lastReturned == null)
                throw new IllegalStateException();

            Node<E> lastNext = lastReturned.next;
            unlink(lastReturned);
            if (next == lastReturned)
                next = lastNext;
            else
                nextIndex--;
            lastReturned = null;
            expectedModCount++;
        }

        public void set(E e) {
            if (lastReturned == null)
                throw new IllegalStateException();
            checkForComodification();
            lastReturned.item = e;
        }

        public void add(E e) {
            checkForComodification();
            lastReturned = null;
            if (next == null)
                linkLast(e);
            else
                linkBefore(e, next);
            nextIndex++;
            expectedModCount++;
        }

        public void forEachRemaining(Consumer<? super E> action) {
            Objects.requireNonNull(action);
            while (modCount == expectedModCount && nextIndex < size) {
                action.accept(next.item);
                lastReturned = next;
                next = next.next;
                nextIndex++;
            }
            checkForComodification();
        }

        final void checkForComodification() {
            if (modCount != expectedModCount)
                throw new ConcurrentModificationException();
        }
    }