【大顶堆+小顶堆】数据流中的中位数

求解代码

private PriorityQueue<Integer> maxHeap = new PriorityQueue<>((a,
            b)->(b - a)); //大顶堆存较小的一半
    private PriorityQueue<Integer> minHeap = new PriorityQueue<>((a, b)->(a - b));

    public void Insert(Integer num) {
        if (maxHeap.isEmpty() || num <= maxHeap.peek()) {
            maxHeap.add(num);
        } else {
            minHeap.add(num);
        }
        balance();
    }

    public Double GetMedian() {
        if (maxHeap.size() == minHeap.size()) {
            return (double)((maxHeap.peek() + minHeap.peek()) / 2.0);
        } else {
            return (double)(maxHeap.size() > minHeap.size() ? maxHeap.peek() :
                            minHeap.peek());
        }
    }

    private void balance() {
        if (Math.abs(maxHeap.size() - minHeap.size()) == 2) {
            if (maxHeap.size() > minHeap.size()) {
                minHeap.add(maxHeap.poll());
            } else {
                maxHeap.add(minHeap.poll());
            }
        }
    }

小贴士

1.Java的PriorityQueue默认是小顶堆，这里采用大顶堆maxHeap存储较小的一半，用小顶堆存储较大的一半。

2.(maxHeap.peek() + minHeap.peek()) / 2.0，除以2.0 触发浮点运算，不会丢失小数位