5.LinkedList

LinkedList即底层基于链表的线性表

当业务中查询少，但需要频繁增，删的场合可使用LinkedList替代ArrayList。

特点：

1. 底层基于链表，不存在初始化容器大小，也不用自动扩容
2. 新增/删除元素平均时间复杂度： O(n)
3. 查找元素平均时间复杂度：O(n)
4. LinkedList除了作为普通的线性链表，还可以作为Java标准的栈和队列实现。所以它的接口要比ArrayList多很多。

LinkedList主要方法

作为普通链表

1. 初始化
   通过new LinkedList() 或 new LinkedList(Collection c)初始化一个链表

2. add(int index, E e)
   在指定索引位新增一个元素，先找到index位的结点，然后将新结点插入到指定位置即可

3. get(int index)
   返回指定索引位的元素，实现也很简单，即从链表的头开始遍历到指定的索引位即可。 相比ArrayList的实现，它的时间复杂度比较高是O(n)

4. remove(int index)
   删除指定索引位的元素，相比ArrayList实现比较简单，不需要移位

作为栈（stack）

0. 初始化空栈

new LinkedList()
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1. 压栈(push)

push(E e)
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2. 出栈(pop)

pop()
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4. 取栈顶元素(peek)

peek()
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作为队列(Queue)

1. 入列(offer)

boolean offer(E e);
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元素入列到队尾

2. 出列(poll)

E poll();
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元素从队首出列

3. 取队首元素（peek）

E peek();
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作为双端队列(Deque)

1. 队首入列(offerFirst)

boolean offerFirst(E e);
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2. 队尾入列(offerLast)

boolean offerLast(E e);
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3. 队首出列(pollFirst)

E pollFirst();
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4. 队尾出列(pollLast)

E pollLast();
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5. 取队首元素(peekFirst)

E peekFirst();
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6. 取队尾元素(peekLast)

E peekLast();
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源码分析

1. add(int index, E element); E remove(int index)

    /**
     * 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) {
    	// 检查index范围， 0 <= index <= size
        checkPositionIndex(index);
		
		// 元素插入到末尾
        if (index == size)
            linkLast(element);
        else // 非末尾，插入到index所在元素的前面
            linkBefore(element, node(index));
    }

    /**
     * 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}
	 * 移除操作非常简单，断开目标元素的上下指针即可
     */
    public E remove(int index) {
        checkElementIndex(index);
        return unlink(node(index));
    }
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2. push, pop, peek

/**
     * Pushes an element onto the stack represented by this list.  In other
     * words, inserts the element at the front of this list.
     *
     * 
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.
     *
     * 
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();
    }

    /**
     * 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;
    }

栈的三个操作非常简单，就是只从头这一端操作。 
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3. offer, poll, peek

/**
     * Adds the specified element as the tail (last element) of this list.
     * 入列操作即将元素加入到队尾
     * @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);
    }
    
    /**
     * Retrieves and removes the head (first element) of this list.
     * 出列操作即从队首取出元素
     * @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, but does not remove, the head (first element) of this list.
     * 获取队首元素，即取首元素
     * @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;
    }


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4. offerFirst, offerLast, pollFirst, pollLast, peekFirst, peekLast
   双端队列的操作

/**
     * 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);
    }

// 双端队列的操作非常简单，队首和队尾同时支持入列，出列和取元素
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