概要
java.lang.Object
java.util.AbstractCollection<E>
java.util.ArrayDeque<E>
public class ArrayDeque<E>
extends AbstractCollection<E>
implements Deque<E>, Cloneable, Serializable
- 对
Deque 接口的实现 - 可调整大小
- 非线程安全
- 作为栈比
Stack 快，作为队列比 LinkedList 快。 - 除了
remove 相关的几个操作是线性时间，其它大部分操作均摊时间都是常数级。
实现
private transient E[] elements;
private transient int head;
private transient int tail;
数据用数组保存，head,tail 指向队列头尾。
static int allocateElements(int numElements) {
int initialCapacity = 8;
// Find the best power of two to hold elements.
// Tests "<=" because arrays aren't kept full.
if (numElements >= initialCapacity) {
initialCapacity = numElements;
initialCapacity |= (initialCapacity >>> 1);
initialCapacity |= (initialCapacity >>> 2);
initialCapacity |= (initialCapacity >>> 4);
initialCapacity |= (initialCapacity >>> 8);
initialCapacity |= (initialCapacity >>> 16);
initialCapacity++;
if (initialCapacity < 0) // Too many elements, must back off
initialCapacity >>>= 1;// Good luck allocating 2 ^ 30 elements
}
return initialCapacity;
}
就像注释里说的那样，这个函数的目的是找到一个刚好大于 numbElements 的数，它是2的n次幂。这是通过移位及或运算的方式实现。
例如这个数为二进制的 abcdefg，a 是最高的不为0的位，即 a 为 1，那么第一次运算后，b 所在位一定为 1，第二次运算后，cd 所在位一定为 1，依此类推。最后全为1,再加1进位。
private void doubleCapacity() {
assert head == tail;
int p = head;
int n = elements.length;
int r = n - p; // number of elements to the right of p
int newCapacity = n << 1;
if (newCapacity < 0)
throw new IllegalStateException("Sorry, deque too big");
Object[] a = new Object[newCapacity];
System.arraycopy(elements, p, a, 0, r);
System.arraycopy(elements, 0, a, r, p);
elements = (E[])a;
head = 0;
tail = n;
}
展示了如何对数组进行扩充，由于这是一个双端队列，head 可能在队列中间，队列现在呈现环状。队列容量增加后，不能简单地直接复制过去，而应该把 head 指向的元素放在新队列数组的 0 号位置，其它元素依次向后排列。注意两次System.arraycopy 分别复制了后一半，和前一半的元素。
public void addFirst(E e) {
if (e == null)
throw new NullPointerException();
elements[head = (head - 1) & (elements.length - 1)] = e;
if (head == tail)
doubleCapacity();
}
首先，参数不能为空，其次， 注意 addFirst 如何巧妙地把 head 的位置后退一位，如果 head 为0的话，& 会使得 head 变成数组最后一个位置。与循环队列概念一致。
public void addLast(E e) {
if (e == null)
throw new NullPointerException();
elements[tail] = e;
if ( (tail = (tail + 1) & (elements.length - 1)) == head)
doubleCapacity();
}
注意新 tail 是如何计算的。通过位运算，比通过 % 显得高大上多了。
public boolean removeFirstOccurrence(Object o) {
if (o == null)
return false;
int mask = elements.length - 1;
int i = head;
E x;
while ( (x = elements[i]) != null) {
if (o.equals(x)) {
delete(i);
return true;
}
i = (i + 1) & mask;
}
return false;
}
初看起来，当队列满时，并且找不到元素时，似乎会永远循环下去，但是实际上，队列永远不会满，所以一定会遇到 null 元素。不会满的原因可以看下之前的 addFirst，当队列将要满的时候，就会自动扩充。任何一个操作之后，tail 指向的位置都为空。
private void checkInvariants() {
assert elements[tail] == null;
assert head == tail ? elements[head] == null :
(elements[head] != null &&
elements[(tail - 1) & (elements.length - 1)] != null);
assert elements[(head - 1) & (elements.length - 1)] == null;
}
private boolean delete(int i) {
checkInvariants();
final E[] elements = this.elements;
final int mask = elements.length - 1;
final int h = head;
final int t = tail;
final int front = (i - h) & mask;
final int back = (t - i) & mask;
// Invariant: head <= i < tail mod circularity
if (front >= ((t - h) & mask))
throw new ConcurrentModificationException();
// Optimize for least element motion
if (front < back) {
if (h <= i) {
System.arraycopy(elements, h, elements, h + 1, front);
} else { // Wrap around
System.arraycopy(elements, 0, elements, 1, i);
elements[0] = elements[mask];
System.arraycopy(elements, h, elements, h + 1, mask - h);
}
elements[h] = null;
head = (h + 1) & mask;
return false;
} else {
if (i < t) { // Copy the null tail as well
System.arraycopy(elements, i + 1, elements, i, back);
tail = t - 1;
} else { // Wrap around
System.arraycopy(elements, i + 1, elements, i, mask - i);
elements[mask] = elements[0];
System.arraycopy(elements, 1, elements, 0, t);
tail = (t - 1) & mask;
}
return true;
}
}
首先注意 checkInvariants 的使用，它保证了执行函数前应该满足的条件，这里是一个前断言。front 和 back 分别代表了要删除的结点距离双端队列头和尾的距离。通过比较 front 和 back 的大小，决定如何移动元素，使得移动次数最少。
如果front<back，还需要区分 h
<= i，决定移动哪一部分。注意这是一个成环的双端队列，画一下位置就明白这些操作了。主要是保证数据是连续的，head/tail 仍然保证其语义。
Deque 有两个 iterator，分别从队列头向后遍历，从队列尾向前遍历。
private class DeqIterator implements Iterator<E> {
/**
* Index of element to be returned by subsequent call to next.
*/
private int cursor = head;
/**
* Tail recorded at construction (also in remove), to stop
* iterator and also to check for comodification.
*/
private int fence = tail;
/**
* Index of element returned by most recent call to next.
* Reset to -1 if element is deleted by a call to remove.
*/
private int lastRet = -1;
public boolean hasNext() {
return cursor != fence;
}
public E next() {
if (cursor == fence)
throw new NoSuchElementException();
E result = elements[cursor];
// This check doesn't catch all possible comodifications,
// but does catch the ones that corrupt traversal
if (tail != fence || result == null)
throw new ConcurrentModificationException();
lastRet = cursor;
cursor = (cursor + 1) & (elements.length - 1);
return result;
}
public void remove() {
if (lastRet < 0)
throw new IllegalStateException();
if (delete(lastRet)) { // if left-shifted, undo increment in next()
cursor = (cursor - 1) & (elements.length - 1);
fence = tail;
}
lastRet = -1;
}
}
这里没有什么特殊的地方，就是通过 cursor 不断指向下一个位置，一直到达尾部。
private class DescendingIterator implements Iterator<E> {
/*
* This class is nearly a mirror-image of DeqIterator, using
* tail instead of head for initial cursor, and head instead of
* tail for fence.
*/
private int cursor = tail;
private int fence = head;
private int lastRet = -1;
public boolean hasNext() {
return cursor != fence;
}
public E next() {
if (cursor == fence)
throw new NoSuchElementException();
cursor = (cursor - 1) & (elements.length - 1);
E result = elements[cursor];
if (head != fence || result == null)
throw new ConcurrentModificationException();
lastRet = cursor;
return result;
}
public void remove() {
if (lastRet < 0)
throw new IllegalStateException();
if (!delete(lastRet)) {
cursor = (cursor + 1) & (elements.length - 1);
fence = head;
}
lastRet = -1;
}
}
这是从后向前的版本。
public void clear() {
int h = head;
int t = tail;
if (h != t) { // clear all cells
head = tail = 0;
int i = h;
int mask = elements.length - 1;
do {
elements[i] = null;
i = (i + 1) & mask;
} while (i != t);
}
}
clear 操作并不是直接将 head/tail 置0就好了，还需要把每个元素置为 null。
public <T> T[] toArray(T[] a) {
int size = size();
if (a.length < size)
a = (T[])java.lang.reflect.Array.newInstance(
a.getClass().getComponentType(), size);
copyElements(a);
if (a.length > size)
a[size] = null;
return a;
}
使用反射机制，生成与泛型一致的数组，再把现有元素复制过去。
