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Android 开发也要懂得数据结构 - LinkList源码

Android 开发也要懂得数据结构 - LinkList源码

作者: 进击的包籽 | 来源:发表于2020-12-17 15:56 被阅读0次

    1.LinkList特点

    • LinkList是双向链表,链表的特点就是插入、删除操作时间复杂度为1,而查找的时间复制度为n。
    • 增删速度相对于数组快,也不需要扩容操作,而查找、修改(需要先查找)速度相对慢一些。
    • Android开发中,如果我们的列表比较多增删的操作,比如即时通讯,新消息过来消息就插入且置顶,左滑或者长按删除之类的操作,就比较适合用LinkList。
    • LinkList也是非线程安全的。

    2.LinkList的继承关系

    • LinkList跟ArrayList相识,也是实现List接口,List接口继承于Collection,Collection继承于Iterable。

    3.LinkList的常用方法

    3.1 构造方法

    • 第一种普通的,不用传参的构造方法,里面也没做什么操作。
        /**
         * Constructs an empty list.
         */
        public LinkedList() {
        }
    
    • 第二种可以传入Collection集合,比如ArrayList传入之后,可以将ArrayList转化为LinkList。
        /**
         * Constructs a list containing the elements of the specified
         * collection, in the order they are returned by the collection's
         * iterator.
         *
         * @param  c the collection whose elements are to be placed into this list
         * @throws NullPointerException if the specified collection is null
         */
        public LinkedList(Collection<? extends E> c) {
            this();
            addAll(c);
        }
    

    3.2 普通增加数据 add(E e)

    • LinkList的每一个元素除了存放内容本身,还有前后指针,分别指向前后节点。因此是双向链表。
        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;
            }
        }
    
    • 普通增加,就是在链表尾部添加数据。
    • 可以看到源码,把 last 链表最后一个节点,先用 l 保存,然后 last 指向 newNode 新节点。
    • 判断如果链表为空,那 first 头结点指向 newNode
    • 否则之前保存的最后一个节点 lnext 后指针指向 newNode
        /**
         * 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++;
        }
    

    3.3 在索引位置插入数据 add(int index, E element)

    • 在索引位置插入数据,不需要像数组那样把后面的元素一个个完后移动,也不需要扩容。
    • 先检查插入的索引位置 index 是否合法。
    • 如果 index位置是尾巴,就直接普通增加数据操作。
    • 否则找到索引位置的节点,不为空哦。
    • 这里看到优化的代码,先进行判断索引位置是否大于一半,进而选择 -- 或者 ++,写这个代码的工程师太细节了
      (╯‵□′)╯︵┻━┻。
        /**
         * Inserts the specified element at the specified position in this list.
         * Shifts the element currently at that position (if any) and any
         * subsequent elements to the right (adds one to their indices).
         *
         * @param index index at which the specified element is to be inserted
         * @param element element to be inserted
         * @throws IndexOutOfBoundsException {@inheritDoc}
         */
        public void add(int index, E element) {
            checkPositionIndex(index);
    
            if (index == size)
                linkLast(element);
            else
                linkBefore(element, node(index));
        }
    
        /**
         * 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;
            }
        }
    
    • 插入操作,就是把索引节点的前节点先用 pred 保存。
    • 创建一个新的节点 newNode,新节点的前指针指向 pred,后指针指向 succ
    • 索引位置的 succ 的前指针指向 newNode,如果之前 succ 的前指针是空,那 newNode 就是头节点,否则 pred 的后指针就是指向 newNode
    • 指针不熟悉的小伙伴可以画图,一下就懂了。
    
       /**
         * Inserts element e before non-null Node succ.
         */
        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);
            succ.prev = newNode;
            if (pred == null)
                first = newNode;
            else
                pred.next = newNode;
            size++;
            modCount++;
        }
    
    • 画图比较易懂


      image.png
      image.png
      image.png
    • 其他增加数据的方法,addFirstaddLastaddAll也是比较简单,跟 add(int index, E element)原理相似。

    3.4 默认删除 remove()

    • 普通的 remove方法移除第一个节点的内容。
    • **element **保存内容,再将引用全部断开,然后再指向下一节点。
       /**
         * Retrieves and removes the head (first element) of this list.
         *
         * @return the head of this list
         * @throws NoSuchElementException if this list is empty
         * @since 1.5
         */
        public E remove() {
            return removeFirst();
        }
    
        /**
         * Removes and returns the first element from this list.
         *
         * @return the first element from this list
         * @throws NoSuchElementException if this list is empty
         */
        public E removeFirst() {
            final Node<E> f = first;
            if (f == null)
                throw new NoSuchElementException();
            return unlinkFirst(f);
        }
    
        /**
         * Unlinks non-null first node f.
         */
        private E unlinkFirst(Node<E> f) {
            // assert f == first && f != null;
           //先拿到存放的内容
            final E element = f.item;
            final Node<E> next = f.next;
            //断开引用,释放内存
            f.item = null;
            f.next = null; // help GC
            first = next;
            if (next == null)
                last = null;
            else
                next.prev = null;
            size--;
            modCount++;
            //返回内容
            return element;
        }
    

    3.5 删除索引位置 remove(int index)

    • 这里的 node(index) 在插入时就分析过,一样样,就是找到指定位置的节点。
    • element 保存着内容,next 为当前节点的后节点,prev 为前节点。
    • 如果前节点为空,那 first 就是 next 节点,否则前节点的后指针指向 next,当前节点的前指针变为 null
    • 如果后节点为空,那 last 就是 prev 节点,否则后节点的前指针指向 prev,当前节点的后指针变为 null
    • 最后返回element
        /**
         * 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));
        }
    
        /**
         * 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;
        }
    

    3.6 删除指定元素 remove(Object o)

    • 遍历链表,找到元素进行删除。
    • LinkList 是可以存放 null 的。
        /**
         * Removes the first occurrence of the specified element from this list,
         * if it is present.  If this list does not contain the element, it is
         * unchanged.  More formally, removes the element with the lowest index
         * {@code i} such that
         * <tt>(o==null&nbsp;?&nbsp;get(i)==null&nbsp;:&nbsp;o.equals(get(i)))</tt>
         * (if such an element exists).  Returns {@code true} if this list
         * contained the specified element (or equivalently, if this list
         * changed as a result of the call).
         *
         * @param o element to be removed from this list, if present
         * @return {@code true} if this list contained the specified element
         */
        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;
        }
    

    3.7 修改元素 set(int index, E element)

    • 找到索引位置的元素,修改内容。
        /**
         * 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;
        }
    

    3.8 长度 size()

    • 返回的是元素的个数。
        /**
         * Returns the number of elements in this list.
         *
         * @return the number of elements in this list
         */
        public int size() {
            return size;
        }
    

    3.9 清空链表 clear()

    • 将每一个节点的内容,前指针,后指针全部置 null
        /**
         * Removes all of the elements from this list.
         * The list will be empty after this call returns.
         */
        public void clear() {
            // Clearing all of the links between nodes is "unnecessary", but:
            // - helps a generational GC if the discarded nodes inhabit
            //   more than one generation
            // - is sure to free memory even if there is a reachable Iterator
            for (Node<E> x = first; x != null; ) {
                Node<E> next = x.next;
                x.item = null;
                x.next = null;
                x.prev = null;
                x = next;
            }
            first = last = null;
            size = 0;
            modCount++;
        }
    

    3.10 找到元素的索引位置 indexOf(Object o)

    • 对比使用的是 Objectequals 方法,所以需要重写对象的 equals
        /**
         * Returns the index of the first occurrence of the specified element
         * in this list, or -1 if this list does not contain the element.
         * More formally, returns the lowest index {@code i} such that
         * <tt>(o==null&nbsp;?&nbsp;get(i)==null&nbsp;:&nbsp;o.equals(get(i)))</tt>,
         * or -1 if there is no such index.
         *
         * @param o element to search for
         * @return the index of the first occurrence of the specified element in
         *         this list, or -1 if this list does not contain the element
         */
        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 {
                for (Node<E> x = first; x != null; x = x.next) {
                    if (o.equals(x.item))
                        return index;
                    index++;
                }
            }
            return -1;
        }
    

    3.11 是否包含 contains(Object o)

    • 是否包含某个元素,用的是 indexOf(Object o) 方法放回是否为 -1,来判断。
        /**
         * Returns {@code true} if this list contains the specified element.
         * More formally, returns {@code true} if and only if this list contains
         * at least one element {@code e} such that
         * <tt>(o==null&nbsp;?&nbsp;e==null&nbsp;:&nbsp;o.equals(e))</tt>.
         *
         * @param o element whose presence in this list is to be tested
         * @return {@code true} if this list contains the specified element
         */
        public boolean contains(Object o) {
            return indexOf(o) != -1;
        }
    

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