5 分钟掌握 JDK 源码之 PriorityQueue

PriorityQueue

初始化

初始容量 11，因此，如果能预估元素个数，最好能设置元素个数，避免元素拷贝

支持从 Collection，SortedSet，PriorityQueue 等初始化

容量扩展

1. 如果容量小于 64，每次扩容为 2 * capacity + 2
2. 如果容量大于等于 64，每次扩容为 capacity + 0.5 * capacity

    // 这里减 8 的原因是，更大的值有的虚拟机会抛 OutOfMemmory
    private static final int MAX_ARRAY_SIZE = Integer.MAX_VALUE - 8;
    
    private void grow(int minCapacity) {
        int oldCapacity = queue.length;
        // Double size if small; else grow by 50%
        int newCapacity = oldCapacity + ((oldCapacity < 64) ?
                                         (oldCapacity + 2) :
                                         (oldCapacity >> 1));
        // overflow-conscious code
        if (newCapacity - MAX_ARRAY_SIZE > 0)
            newCapacity = hugeCapacity(minCapacity);
        queue = Arrays.copyOf(queue, newCapacity);
    }

    private static int hugeCapacity(int minCapacity) {
        if (minCapacity < 0) // overflow
            throw new OutOfMemoryError();
        return (minCapacity > MAX_ARRAY_SIZE) ?
            Integer.MAX_VALUE :
            MAX_ARRAY_SIZE;
    }

实现原理

1. 通过数组模拟堆
2. 核心操作 siftUp, siftDown

父子节点关系

1. 父节点总是小于等于孩子节点
2. 同一层孩子节点间的大小无需维护

parent = (k - 1) >>> 1

left_child = (k << 1) + 1

right_child = (k << 1) + 2

SiftUp

算法复杂度：Log(N)

    private void siftUp(int k, E x) {
        if (comparator != null)
            siftUpUsingComparator(k, x);
        else
            siftUpComparable(k, x);
    }

    @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;
            Object e = queue[parent];
            if (key.compareTo((E) e) >= 0)
                break;
            queue[k] = e;
            k = parent;
        }
        queue[k] = key;
    }

    @SuppressWarnings("unchecked")
    private void siftUpUsingComparator(int k, E x) {
        while (k > 0) {
            int parent = (k - 1) >>> 1;
            Object e = queue[parent];
            if (comparator.compare(x, (E) e) >= 0)
                break;
            queue[k] = e;
            k = parent;
        }
        queue[k] = x;
    }

调整的过程为：从 k 指定的位置开始，将 x 逐层与当前点的 parent 进行比较并交换，直到满足x >= queue[parent] 为止。

SiftDown

算法复杂度：Log(N)

    private void siftDown(int k, E x) {
        if (comparator != null)
            siftDownUsingComparator(k, x);
        else
            siftDownComparable(k, x);
    }

    @SuppressWarnings("unchecked")
    private void siftDownComparable(int k, E x) {
        Comparable<? super E> key = (Comparable<? super E>)x;
        int half = size >>> 1;        // loop while a non-leaf
        while (k < half) {
            int child = (k << 1) + 1; // assume left child is least
            Object c = queue[child];
            int right = child + 1;
            if (right < size &&
                ((Comparable<? super E>) c).compareTo((E) queue[right]) > 0)
                c = queue[child = right];
            if (key.compareTo((E) c) <= 0)
                break;
            queue[k] = c;
            k = child;
        }
        queue[k] = key;
    }

    @SuppressWarnings("unchecked")
    private void siftDownUsingComparator(int k, E x) {
        int half = size >>> 1;
        while (k < half) {
            int child = (k << 1) + 1;
            Object c = queue[child];
            int right = child + 1;
            if (right < size &&
                comparator.compare((E) c, (E) queue[right]) > 0)
                c = queue[child = right];
            if (comparator.compare(x, (E) c) <= 0)
                break;
            queue[k] = c;
            k = child;
        }
        queue[k] = x;
    }

调整的过程为：从 k 指定的位置开始，将 x 逐层向下与当前点的左右孩子中较小的那个交换，直到 x 小于或等于左右孩子中的任何一个为止。

poll

算法复杂度： O(N)

   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;
    }

offer

算法复杂度：Log(N)

   public boolean offer(E e) {
        if (e == null)
            throw new NullPointerException();
        modCount++;
        int i = size;
        if (i >= queue.length)
            grow(i + 1);
        size = i + 1;
        if (i == 0)
            queue[0] = e;
        else
            siftUp(i, e);
        return true;
    }

remove

算法复杂度：Log(N)

   public boolean remove(Object o) {
        int i = indexOf(o);
        if (i == -1)
            return false;
        else {
            removeAt(i);
            return true;
        }
    }   
   
   private E removeAt(int i) {
        // assert i >= 0 && i < size;
        modCount++;
        int s = --size;
        if (s == i) // removed last element
            queue[i] = null;
        else {
            E moved = (E) queue[s];
            queue[s] = null;
            siftDown(i, moved);
            if (queue[i] == moved) {
                siftUp(i, moved);
                if (queue[i] != moved)
                    return moved;
            }
        }
        return null;
    }

总结

基于数组的最小堆实现，整个实现比较精巧，强烈建议掌握，在解决 LeetCode 相关问题是大杀器。

1. 扩容对性能的影响。如果能够预估大小，最好初始化指定容量。
2. offer 和 poll 中的 siftDown 和 siftUp