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愉快地聊一聊ArrayDeque的特点吧~（以下都是基于jdk1.8）
| 一棵树 |
ArrayDeque的继承树如下图：
| 基本特点 |
（1）双端队列，可从两端添加、删除元素。作为队列使用时，性能优于LinkedList。作为栈使用时，性能优于Stack。
（2）底层使用可变数组Object[] elements， 数组容量按需增长
（3）不能存储null
（4）支持双向迭代器遍历
（5）线程不安全
（6）fail-fast机制。
（7）最小数组容量MIN_INITIAL_CAPACITY = 8。Must be a power of 2
（8）默认数组初始容量是16
（9）调用指定初始容量的构造函数，并不会按照指定值分配容量。
（10）先添加，再判断是否需要扩容
| 源码之旅 |
这里只取部分源码进行分析：指定初始容量的构造函数、扩容机制，以及主要方法。
好了，先把类中定义的变量熟悉一下：
/** * The array in which the elements of the deque are stored. * The capacity of the deque is the length of this array, which is * always a power of two. The array is never allowed to become * full, except transiently within an addX method where it is * resized (see doubleCapacity) immediately upon becoming full, * thus avoiding head and tail wrapping around to equal each * other. We also guarantee that all array cells not holding * deque elements are always null. */
transient Object[] elements; // non-private to simplify nested class access
/** * The index of the element at the head of the deque (which is the * element that would be removed by remove() or pop()); or an * arbitrary number equal to tail if the deque is empty. */
transient int head;
/** * The index at which the next element would be added to the tail * of the deque (via addLast(E), add(E), or push(E)). */
transient int tail;
/** * The minimum capacity that we'll use for a newly created deque. * Must be a power of 2. */
private static final int MIN_INITIAL_CAPACITY = 8;
（1）指定初始容量的构造函数：
找到该构造函数，从此入手：
/** * Constructs an empty array deque with an initial capacity * sufficient to hold the specified number of elements. * * @param numElements lower bound on initial capacity of the deque */
public ArrayDeque(int numElements) {
allocateElements(numElements);
}
再看allocateElements方法：
/** * Allocates empty array to hold the given number of elements. * * @param numElements the number of elements to hold */
private void allocateElements(int numElements) {
int initialCapacity = MIN_INITIAL_CAPACITY;
// 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
}
elements = new Object[initialCapacity];
}
拿给定元素数量与数组最小容量8做比较，因为此集合不允许数组变满（添加元素的方法中，数组容量一满就立刻扩容），所以当给定元素数量>=数组最小容量8时，会进行一系列的无符号右移运算、或运算，以便找到能够容纳给定元素的最佳的2的幂次方。
这个最佳的2的幂次方就是调用该构造函数后底层为我们分配的数组容量。
（2）扩容机制：
找到扩容的核心方法：
/** * Doubles the capacity of this deque. Call only when full, i.e., * when head and tail have wrapped around to become equal. */
private void doubleCapacity() {
//断言 判断head与tail指针是否相等
assert head == tail;
int p = head;
int n = elements.length;
int r = n - p; // number of elements to the right of p
//左移1位，相当于*2操作，只是<<效率要高于*运算。
int newCapacity = n << 1;
if (newCapacity < 0)
throw new IllegalStateException("Sorry, deque too big");
//扩容，实际上是定义了一个指定容量的数组，将elements数组中的元素复制到新数组a中。
Object[] a = new Object[newCapacity];
System.arraycopy(elements, p, a, 0, r);
System.arraycopy(elements, 0, a, r, p);
//重新设置head和tail的指针
elements = a;
head = 0;
tail = n;
}
（3）主要方法：
添加元素：
// The main insertion and extraction methods are addFirst,
// addLast, pollFirst, pollLast. The other methods are defined in
// terms of these.
/** * 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();
//此运算可以快速定位到要插入的位置，实际上是从数组最右侧开始插入的，head是递减的
elements[head = (head - 1) & (elements.length - 1)] = e;
//head与tail重叠时，开始扩容
if (head == tail)
doubleCapacity();
}
/** * 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();
//tail初始值是0，指向待插入元素的位置，tail是递增的
elements[tail] = e;
//先插入元素，再判断是否需要扩容。
//tail + 1 & (elements.length - 1 )用于定位下一个待插入元素的位置。
//如果tail与head重叠，数组容量已满，
if ( (tail = (tail + 1) & (elements.length - 1)) == head)
doubleCapacity();
}
/** * Inserts the specified element at the front of this deque. * * @param e the element to add * @return {@code true} (as specified by {@link Deque#offerFirst}) * @throws NullPointerException if the specified element is null */
public boolean offerFirst(E e) {
addFirst(e);
return true;
}
/** * Inserts the specified element at the end of this deque. * * @param e the element to add * @return {@code true} (as specified by {@link Deque#offerLast}) * @throws NullPointerException if the specified element is null */
public boolean offerLast(E e) {
addLast(e);
return true;
}
删除首尾元素：
public E removeFirst() {
E x = pollFirst();
if (x == null)
throw new NoSuchElementException();
return x;
}
public E removeLast() {
E x = pollLast();
if (x == null)
throw new NoSuchElementException();
return x;
}
public E pollFirst() {
int h = head;
@SuppressWarnings("unchecked")
E result = (E) elements[h];
// Element is null if deque empty
if (result == null)
return null;
elements[h] = null; // Must null out slot
head = (h + 1) & (elements.length - 1);
return result;
}
public E pollLast() {
int t = (tail - 1) & (elements.length - 1);
@SuppressWarnings("unchecked")
E result = (E) elements[t];
if (result == null)
return null;
elements[t] = null;
tail = t;
return result;
}
删除指定元素：
/** * Removes the first occurrence of the specified element in this * deque (when traversing the deque from head to tail). * If the deque does not contain the element, it is unchanged. * More formally, removes the first element {@code e} such that * {@code o.equals(e)} (if such an element exists). * Returns {@code true} if this deque contained the specified element * (or equivalently, if this deque changed as a result of the call). * * @param o element to be removed from this deque, if present * @return {@code true} if the deque contained the specified element */
public boolean removeFirstOccurrence(Object o) {
if (o == null)
return false;
int mask = elements.length - 1;
int i = head;
Object x;
while ( (x = elements[i]) != null) {
if (o.equals(x)) {
delete(i);
return true;
}
i = (i + 1) & mask;
}
return false;
}
/** * Removes the last occurrence of the specified element in this * deque (when traversing the deque from head to tail). * If the deque does not contain the element, it is unchanged. * More formally, removes the last element {@code e} such that * {@code o.equals(e)} (if such an element exists). * Returns {@code true} if this deque contained the specified element * (or equivalently, if this deque changed as a result of the call). * * @param o element to be removed from this deque, if present * @return {@code true} if the deque contained the specified element */
public boolean removeLastOccurrence(Object o) {
if (o == null)
return false;
int mask = elements.length - 1;
int i = (tail - 1) & mask;
Object x;
while ( (x = elements[i]) != null) {
if (o.equals(x)) {
delete(i);
return true;
}
i = (i - 1) & mask;
}
return false;
}
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;
}
/** * Removes the element at the specified position in the elements array, * adjusting head and tail as necessary. This can result in motion of * elements backwards or forwards in the array. * * <p>This method is called delete rather than remove to emphasize * that its semantics differ from those of {@link List#remove(int)}. * * @return true if elements moved backwards */
private boolean delete(int i) {
checkInvariants();
final Object[] 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;
}
}
暂时先写到这里了~
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