内容简介:上一篇文章我们介绍了JDK中ArrayList的实现,ArrayList底层结构是一个Object[]数组,通过拷贝,复制等一系列封装的操作,将数组封装为一个几乎是无限的容器。今天我们来介绍JDK中List接口的另外一种实现,基于链表结构的LinkedList。ArrayList由于基于数组,所以在随机访问方面优势比较明显,在删除、插入方面性能会相对偏弱些(当然与删除、插入的位置有很大关系)。那么LinkedList有哪些优势呢?它在删除、插入方面的操作很简单(只是调整相关指针而已)。但是随机访问方面要逊色
上一篇文章我们介绍了JDK中ArrayList的实现,ArrayList底层结构是一个Object[]数组,通过拷贝,复制等一系列封装的操作,将数组封装为一个几乎是无限的容器。今天我们来介绍JDK中List接口的另外一种实现,基于链表结构的LinkedList。ArrayList由于基于数组,所以在随机访问方面优势比较明显,在删除、插入方面性能会相对偏弱些(当然与删除、插入的位置有很大关系)。那么LinkedList有哪些优势呢?它在删除、插入方面的操作很简单(只是调整相关指针而已)。但是随机访问方面要逊色写。下面我们还是从源码上来看下这种链表结构的List。
LinkedList类主要字段
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; 复制代码
我们看到字段非常少,size表示当前节点数量,first指向链表的起始元素、last指向链表的最后一个元素。
Node结构
从上面主要字段看出,LinkedList链表的Item就是一个Node结构,那么Node结构是怎样的呢?源码如下:
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; } } 复制代码
我们看到Node结构包含一个前驱prev指针,item(value)、后继next指针三个部分。结合上面的描述,我们知道了LinkedList的主要结构。如图:
第一个节点的prev指向NULL,最后一个节点的next指向NULL。其余节点通过prev与next串联起来,LinkedList提供了从任意节点都能进行向前或先后遍历的能力。
LinkedList相关方法解析
构造函数
/** * Constructs an empty list. */ public LinkedList() { } /** * 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); } 复制代码
由于LinkedList是通过prev与next指针链接起来的,有元素添加时只需要一个个设置指针将其链接起来即可,所以构造函数相对较简洁。我们重点来看下第二个构造函数中的addAll方法。
addAll方法
/** * Appends all of the elements in the specified collection to the end of * this list, in the order that they are returned by the specified * collection's iterator. The behavior of this operation is undefined if * the specified collection is modified while the operation is in * progress. (Note that this will occur if the specified collection is * this list, and it's nonempty.) * * @param c collection containing elements to be added to this list * @return {@code true} if this list changed as a result of the call * @throws NullPointerException if the specified collection is null */ public boolean addAll(Collection<? extends E> c) { return addAll(size, c); } /** * Inserts all of the elements in the specified collection into this * list, starting at the specified position. Shifts the element * currently at that position (if any) and any subsequent elements to * the right (increases their indices). The new elements will appear * in the list in the order that they are returned by the * specified collection's iterator.' * * @param index index at which to insert the first element * from the specified collection * @param c collection containing elements to be added to this list * @return {@code true} if this list changed as a result of the call * @throws IndexOutOfBoundsException {@inheritDoc} * @throws NullPointerException if the specified collection is null */ 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; } /** * 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; } } 复制代码
addAll方法是将Collection集合插入链表。下面我们来仔细分析整个过程(涉及比较多的指针操作)。
- 首先代码检查index的值的正确性,如果index位置不合理会直接抛出异常。
- 然后将待插入集合转化成数组,判断集合长度。
- 根据index值,分别设置pred和succ指针。如果插入的位置是当前链表尾部,那么pred指向最后一个元素,succ暂时设置为NULL即可。如果插入位置在链表中间,那么先通过node方法找到当前链表的index位置的元素,succ指向它。pred指向待插入位置的前一个节点,succ指向当前index位置的节点,新插入的节点就是在pred和succ节点之间。
- for循环创建Node节点,先将pred.next指向新创建的节点,然后pred指向后移,指向新创建的Node节点,重复上述过程,这样一个个节点就被创建,链接起来了。
- 最后根据情况不同,将succ指向的那个节点作为最后的节点,当然如果succ为NULL的话,last指针指向pred。
removeFirst()方法和removeLast()方法
removeFirst方法会返回当前链表的头部节点值,然后将头结点指向下一个节点,我们通过源码来分析:
/** * 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; final E element = f.item; final Node<E> next = f.next; f.item = null; f.next = null; // help GC first = next; if (next == null) last = null; else next.prev = null; size--; modCount++; return element; } 复制代码
我们看到主要逻辑在unlinkFirst方法中,逻辑还是比较清晰的,first指针指向next节点,该节点作为新的链表头部,只是最后需要处理下边界值(next==null)的情况。removeLast方法类似,大家可以去分析源码。
addFirst方法和addLast()方法
addFirst方法是将新节点插入链表,并且将新节点作为链表头部,下面我们来看源码:
/** * Inserts the specified element at the beginning of this list. * * @param e the element to add */ public void addFirst(E e) { linkFirst(e); } /** * Links e as first element. */ 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++; } 复制代码
代码逻辑比较清晰,newNode节点在创建时,由于是作为新的头结点的,所以prev必须是NULL的,next是指向当前头结点f。接下来就是设置first,处理边界值了。
下面我们来看下addLast方法,源码如下:
/** * Appends the specified element to the end of this list. * * <p>This method is equivalent to {@link #add}. * * @param e the element to add */ public void addLast(E e) { linkLast(e); } /** * 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++; } 复制代码
add方法和remove方法
这两个方法是我们使用频率很高的方法,我们来看下其内部实现:
/** * 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; } /** * 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 ? get(i)==null : 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; } /** * 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; } 复制代码
我们看到add方法其实就是对linkLast方法的封装(当然,这是末尾添加)。remove方法逻辑会复杂些,需要先找到指定节点,然后调用unlink方法。
unlink方法解析
我们看到unlink方法首先将需要删除的节点的prev和next保存起来,因为后面需要将两者连接起来。然后将prev和next分别判断设置(包括边界值的考虑),最后将x节点的数据设置为NULL。
clear方法
LinkedList链表结构的,它的clear方法是如何实现的呢?我们来看下:
/** * 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++; } 复制代码
代码还是比较清晰的,就是从头结点开始,将Node节点一个个的设置为NULL,方便GC回收。
LinkedList与队列操作
有数据结构基础的同学应该都知道队列的结构,这是一种先进先出的结构。从JDK1.5开始,LinkedList内部集成了队列的操作,LinkedList可以当做一个基本的队列进行使用。下面我们从队列的角度来看下LinkedList提供的相关方法。
peek、poll、element、remove方法
/** * Retrieves, but does not remove, the head (first element) of this list. * * @return the head of this list, or {@code null} if this list is empty * @since 1.5 */ public E peek() { final Node<E> f = first; return (f == null) ? null : f.item; } /** * Retrieves, but does not remove, 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 element() { return getFirst(); } /** * Retrieves and removes the head (first element) of this list. * * @return the head of this list, or {@code null} if this list is empty * @since 1.5 */ public E poll() { final Node<E> f = first; return (f == null) ? null : unlinkFirst(f); } /** * 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(); } 复制代码
从上面的方法,我们知道peek、element方法只返回队列头部数据,不移除头部。而poll、remove方法返回队列头部数据的同是,还会移除头部。
offer方法
/** * Adds the specified element as the tail (last element) of this list. * * @param e the element to add * @return {@code true} (as specified by {@link Queue#offer}) * @since 1.5 */ public boolean offer(E e) { return add(e); } 复制代码
从上面的代码中我们看到,offer方法其实就是入队操作。
LinkedList与双端队列
上面我们介绍了使用LinkedList来作为队列的相关方法,在JDK6中添加相关方法让LinkedList支持双端队列。源代码如下:
// Deque operations /** * Inserts the specified element at the front of this list. * * @param e the element to insert * @return {@code true} (as specified by {@link Deque#offerFirst}) * @since 1.6 */ public boolean offerFirst(E e) { addFirst(e); return true; } /** * Inserts the specified element at the end of this list. * * @param e the element to insert * @return {@code true} (as specified by {@link Deque#offerLast}) * @since 1.6 */ public boolean offerLast(E e) { addLast(e); return true; } /** * Retrieves, but does not remove, the first element of this list, * or returns {@code null} if this list is empty. * * @return the first element of this list, or {@code null} * if this list is empty * @since 1.6 */ public E peekFirst() { final Node<E> f = first; return (f == null) ? null : f.item; } /** * Retrieves, but does not remove, the last element of this list, * or returns {@code null} if this list is empty. * * @return the last element of this list, or {@code null} * if this list is empty * @since 1.6 */ public E peekLast() { final Node<E> l = last; return (l == null) ? null : l.item; } /** * Retrieves and removes the first element of this list, * or returns {@code null} if this list is empty. * * @return the first element of this list, or {@code null} if * this list is empty * @since 1.6 */ public E pollFirst() { final Node<E> f = first; return (f == null) ? null : unlinkFirst(f); } /** * Retrieves and removes the last element of this list, * or returns {@code null} if this list is empty. * * @return the last element of this list, or {@code null} if * this list is empty * @since 1.6 */ public E pollLast() { final Node<E> l = last; return (l == null) ? null : unlinkLast(l); } 复制代码
上面的代码逻辑比较清楚,就不详细介绍了。
LinkedList与栈(Stack)
堆栈大伙肯定很熟悉,是一种先进后出的结构。类似于叠盘子,一般我们使用的时候肯定从最上面拿取。栈也是这样,最后进入的,最先出去。LinkedList在JDK6的时候也添加了对栈的支持。我们来看相关源码:
/** * Pushes an element onto the stack represented by this list. In other * words, inserts the element at the front of this list. * * <p>This method is equivalent to {@link #addFirst}. * * @param e the element to push * @since 1.6 */ public void push(E e) { addFirst(e); } /** * Pops an element from the stack represented by this list. In other * words, removes and returns the first element of this list. * * <p>This method is equivalent to {@link #removeFirst()}. * * @return the element at the front of this list (which is the top * of the stack represented by this list) * @throws NoSuchElementException if this list is empty * @since 1.6 */ public E pop() { return removeFirst(); } 复制代码
我们看到LinkedList封装的push和pop操作其实就是对first头结点的操作。通过对头结点不短了的push、pop来模拟堆栈先进后出的结构。
LinkedList与迭代器
private 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(); } } 复制代码
从上面的迭代器的源码我们可以知道以下几点:
- 1、LinkedList通过自定义迭代器实现了往前往后两个方向的遍历。
- 2、remove方法中next == lastReturned条件的判断是针对上一次对链表进行了previous操作后进行的判断。因为上一次previous操作后next指针会“悬空”。需要将其设置为next节点。
LinkedList遍历相关问题
对于集合来说,遍历是非常常规的操作。但是对于LinkedList来说,遍历的时候需要选择合适的方法,因为不合理的方法对于性能有非常大的差别。我们通过例子来看:
List<String> list=new LinkedList<>(); for(int i=0;i<10000;i++) { list.add(String.valueOf(i)); } //遍历方法一 long time=System.currentTimeMillis(); for(int i=0;i<list.size();i++) { list.get(i); } System.out.println(System.currentTimeMillis()-time); time=System.currentTimeMillis(); Iterator<String> iterator=list.iterator(); while (iterator.hasNext()) { iterator.next(); } iterator.remove(); System.out.println(System.currentTimeMillis()-time); 复制代码
输出如下:
size:10000的情况 120 2 size:100000的情况 28949 2 复制代码
同样是遍历方法,为什么性能差别几十倍,设置上万倍呢?研究过源码的同学应该能发现其中的奥秘。我们来看get方法的逻辑:
/** * 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; } } 复制代码
我们看到,我们get(index)的时候,都需要从头,或者从尾部慢慢循环过来。get(4000)的时候需要从0-4000进行遍历。get(4001)的时候还是需要从0-4001进行遍历。做了无数的无用功。但是迭代器就不一样了。迭代器通过next指针,能指向下一个节点,无需做额外的遍历,速度非常快。
总结
- 1、LinkedList在添加及修改时候效率较高,只需要设置前后节点即可(ArrayList还需要拷贝前后数据)。
- 2、LinkedList不同的遍历性能差距极大,推荐使用迭代器进行遍历。LinkedList在随机访问方面性能一般(ArrayList随机方法可以使用基地址+偏移量的方式访问)
- LinkedList提供作为队列、堆栈的相关方法。
以上所述就是小编给大家介绍的《LinkedList源码解析》,希望对大家有所帮助,如果大家有任何疑问请给我留言,小编会及时回复大家的。在此也非常感谢大家对 码农网 的支持!
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