java.util.TaskQueue的最小堆排序算法的应用
其中fixup和fixdown就是堆排序的使用。
/** * This class represents a timer task queue: a priority queue of TimerTasks, * ordered on nextExecutionTime. Each Timer object has one of these, which it * shares with its TimerThread. Internally this class uses a heap, which * offers log(n) performance for the add, removeMin and rescheduleMin * operations, and constant time performance for the getMin operation. */ class TaskQueue { /** * Priority queue represented as a balanced binary heap: the two children * of queue[n] are queue[2*n] and queue[2*n+1]. The priority queue is * ordered on the nextExecutionTime field: The TimerTask with the lowest * nextExecutionTime is in queue[1] (assuming the queue is nonempty). For * each node n in the heap, and each descendant of n, d, * n.nextExecutionTime <= d.nextExecutionTime. */ private TimerTask[] queue = new TimerTask[128]; /** * The number of tasks in the priority queue. (The tasks are stored in * queue[1] up to queue[size]). */ private int size = 0; /** * Returns the number of tasks currently on the queue. */ int size() { return size; } /** * Adds a new task to the priority queue. */ void add(TimerTask task) { // Grow backing store if necessary if (size + 1 == queue.length) queue = Arrays.copyOf(queue, 2*queue.length); queue[++size] = task; fixUp(size); } /** * Return the "head task" of the priority queue. (The head task is an * task with the lowest nextExecutionTime.) */ TimerTask getMin() { return queue[1]; } /** * Return the ith task in the priority queue, where i ranges from 1 (the * head task, which is returned by getMin) to the number of tasks on the * queue, inclusive. */ TimerTask get(int i) { return queue[i]; } /** * Remove the head task from the priority queue. */ void removeMin() { queue[1] = queue[size]; queue[size--] = null; // Drop extra reference to prevent memory leak fixDown(1); } /** * Removes the ith element from queue without regard for maintaining * the heap invariant. Recall that queue is one-based, so * 1 <= i <= size. */ void quickRemove(int i) { assert i <= size; queue[i] = queue[size]; queue[size--] = null; // Drop extra ref to prevent memory leak } /** * Sets the nextExecutionTime associated with the head task to the * specified value, and adjusts priority queue accordingly. */ void rescheduleMin(long newTime) { queue[1].nextExecutionTime = newTime; fixDown(1); } /** * Returns true if the priority queue contains no elements. */ boolean isEmpty() { return size==0; } /** * Removes all elements from the priority queue. */ void clear() { // Null out task references to prevent memory leak for (int i=1; i<=size; i++) queue[i] = null; size = 0; } /** * Establishes the heap invariant (described above) assuming the heap * satisfies the invariant except possibly for the leaf-node indexed by k * (which may have a nextExecutionTime less than its parent's). * * This method functions by "promoting" queue[k] up the hierarchy * (by swapping it with its parent) repeatedly until queue[k]'s * nextExecutionTime is greater than or equal to that of its parent. */ private void fixUp(int k) { while (k > 1) { int j = k >> 1; if (queue[j].nextExecutionTime <= queue[k].nextExecutionTime) break; TimerTask tmp = queue[j]; queue[j] = queue[k]; queue[k] = tmp; k = j; } } /** * Establishes the heap invariant (described above) in the subtree * rooted at k, which is assumed to satisfy the heap invariant except * possibly for node k itself (which may have a nextExecutionTime greater * than its children's). * * This method functions by "demoting" queue[k] down the hierarchy * (by swapping it with its smaller child) repeatedly until queue[k]'s * nextExecutionTime is less than or equal to those of its children. */ private void fixDown(int k) { int j; while ((j = k << 1) <= size && j > 0) { if (j < size && queue[j].nextExecutionTime > queue[j+1].nextExecutionTime) j++; // j indexes smallest kid if (queue[k].nextExecutionTime <= queue[j].nextExecutionTime) break; TimerTask tmp = queue[j]; queue[j] = queue[k]; queue[k] = tmp; k = j; } } /** * Establishes the heap invariant (described above) in the entire tree, * assuming nothing about the order of the elements prior to the call. */ void heapify() { for (int i = size/2; i >= 1; i--) fixDown(i); } }