一 概览
image.png
由上图可知，Queue分成只扩展AbstractQueue的，实现BlockingQueue接口的，实现TransferQueue的，实现BlockingDeque的以及实现Deque接口的五类
二 PriorityQueue
- siftUp
/**
* Inserts item x at position k, maintaining heap invariant by
* promoting x up the tree until it is greater than or equal to
* its parent, or is the root.
*
* To simplify and speed up coercions and comparisons. the
* Comparable and Comparator versions are separated into different
* methods that are otherwise identical. (Similarly for siftDown.)
*
* @param k the position to fill
* @param x the item to insert
*/
@SuppressWarnings("unchecked")
private void siftUpComparable(int k, E x) {
Comparable<? super E> key = (Comparable<? super E>) x;
while (k > 0) {
int parent = (k - 1) >>> 1; //不像siftDown那样需要考虑找左右 孩子中的最小节点，siftUp只需要比较key跟parent节点即可
Object e = queue[parent];
if (key.compareTo((E) e) >= 0)
break;
queue[k] = e;
k = parent;
}
queue[k] = key;
}
- siftDown
/**
* Inserts item x at position k, maintaining heap invariant by
* demoting x down the tree repeatedly until it is less than or
* equal to its children or is a leaf.
*
* @param k the position to fill
* @param x the item to insert
*/
private void siftDownComparable(int k, E x) {
Comparable<? super E> key = (Comparable<? super E>)x;
int half = size >>> 1; // loop while a non-leaf 先看看是不是非叶子节点
while (k < half) {
int child = (k << 1) + 1; // assume left child is least 先假设左孩子节点最小
Object c = queue[child];
int right = child + 1;
if (right < size &&
((Comparable<? super E>) c).compareTo((E) queue[right]) > 0)
c = queue[child = right]; //如果右孩子更小，则将right赋给child，把queue[right]赋给c
if (key.compareTo((E) c) <= 0) //如果目标节点值小于等于c，则break，否则，继续循环
break;
queue[k] = c;
k = child;
}
queue[k] = key;
}
- offer
public boolean offer(E e) {
if (e == null)
throw new NullPointerException();
modCount++;
int i = size;
if (i >= queue.length)
grow(i + 1); //扩容，与map和list的grow类似
size = i + 1;
if (i == 0)
queue[0] = e;
else
siftUp(i, e);
return true;
}
offer(E e)方法的实现思路如下：
1）首先检查要添加的元素e是否为null，如果为null，则抛空指针异常，如果不为null，则进行2
2）判断数组是否已满，如果已满，则进行扩容，否则将元素加入到数组末尾即可。但是由于这个数组表示的是一个“二叉堆”，因此还需要进行相应的调整操作，使得这个数组在添加元素之后依然保持的是二叉堆的特性。
- poll
public E poll() {
if (size == 0)
return null;
int s = --size;
modCount++;
E result = (E) queue[0];
E x = (E) queue[s];
queue[s] = null;
if (s != 0)
siftDown(0, x);
return result;
}
poll 方法的思想为：取出 queue[0] 元素，然后将 queue[size-1] 插入到 queue[0] , 然后向下沉来保持二叉堆的特性。
