Java Deque

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关于Deque，之前我的总结是，如果要当作Stack使用，则stick to Stack methods，比如push, pop, peek。如果当作Queue使用，则stick to Queue method，比如add/offer, poll/remove, peek。在实现上，我一般使用的是ArrayDeque, a resizable double-ended array。

今天突然想到一个问题，用Deque实现的Queue或者Stack，在使用enhanced for loop的时候，Java是怎么知道元素弹出的正确顺序呢? 或者如果混用Queue和Stack的方法，peek会弹出什么结果？iterator会给出什么顺序的结果呢？

我们来看看ArrayDeque的源码:

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transient int head;
transient int tail;

这里有2个pointers, head和tail，当arraydeque是empty的时候，head和tail重叠。

对于push来说，移动的是head，head的值减1再使用，并且用的是modulus circularly decrement。

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/**
* Pushes an element onto the stack represented by this deque.  In other
* words, inserts the element at the front of this deque.
*
* <p>This method is equivalent to {@link #addFirst}.
*
* @param e the element to push
* @throws NullPointerException if the specified element is null
*/
public void push(E e) 
{
    addFirst(e);
}

/**
* Inserts the specified element at the front of this deque.
*
* @param e the element to add
* @throws NullPointerException if the specified element is null
*/
public void addFirst(E e) 
{
    if (e == null)
        throw new NullPointerException();
    final Object[] es = elements;
    es[head = dec(head, es.length)] = e;
    if (head == tail)
        grow(1);
}

/**
* Circularly decrements i, mod modulus.
* Precondition and postcondition: 0 <= i < modulus.
*/
static final int dec(int i, int modulus) 
{
    if (--i < 0) i = modulus - 1;
    return i;
}

对于add/offer来说，tail的值用了再加1，并且用的是modulus circularl increment。

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/**
* Inserts the specified element at the end of this deque.
*
* <p>This method is equivalent to {@link #addLast}.
*
* @param e the element to add
* @return {@code true} (as specified by {@link Collection#add})
* @throws NullPointerException if the specified element is null
*/
public boolean add(E e) 
{
    addLast(e);
    return true;
}

/**
* Inserts the specified element at the end of this deque.
*
* <p>This method is equivalent to {@link #add}.
*
* @param e the element to add
* @throws NullPointerException if the specified element is null
*/
public void addLast(E e) 
{
    if (e == null)
        throw new NullPointerException();
    final Object[] es = elements;
    es[tail] = e;
    if (head == (tail = inc(tail, es.length)))
        grow(1);
}

/**
* Circularly increments i, mod modulus.
* Precondition and postcondition: 0 <= i < modulus.
*/
static final int inc(int i, int modulus) 
{
    if (++i >= modulus) i = 0;
    return i;
}

peek总是从head pointer取值

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/**
* Retrieves, but does not remove, the head of the queue represented by
* this deque, or returns {@code null} if this deque is empty.
*
* <p>This method is equivalent to {@link #peekFirst}.
*
* @return the head of the queue represented by this deque, or
*         {@code null} if this deque is empty
*/
public E peek() 
{
    return peekFirst();
}

iterator总是从head pointer -> tail pointer

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/**
* Returns an iterator over the elements in this deque.  The elements
* will be ordered from first (head) to last (tail).  This is the same
* order that elements would be dequeued (via successive calls to
* {@link #remove} or popped (via successive calls to {@link #pop}).
*
* @return an iterator over the elements in this deque
*/
public Iterator<E> iterator() 
{
    return new DeqIterator();
}

所以现在情形就很清楚了，来看一个例子。首先按照顺序push [1 2 3], 这时peek是3，然后再add/offer [4 5 6], 这时peek还是3，然后iterator的结果是: [3 2 1 4 5 6]

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Deque<Integer> dq = new ArrayDeque<>();
dq.push(1);
dq.push(2);
dq.push(3);
// now peek is 3
dq.add(4);
dq.add(5);
dq.add(6);
// now peek is still 3
for (int ele: dq)
{
    System.out.println(ele);
}
// 3 2 1 4 5 6

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