在线求中位数

在线求第k个数做得多了，在线求中位数也是用堆，一个最大堆，一个最小堆。

思想大概是这样子的:

1. 一个最大堆，一个最小堆，最大堆对应于前n/(n+1)个数，最小堆对应于后n/n+1个数；假设最大堆堆项元素为n1， 最小堆堆顶为n2， 则n1 <= n2;
2. 确保两个堆的大小最多只差1. 设最大堆大小为s1， 最小堆大小为s2，则abs(s1-s2) <= 1；
3. 对于新来的数m，分情况调整：
   1. 如果s1== s2, 那么：如果m<= n2, m插入到最大堆，s1= s1+1; 否则插入到最小堆，s2=s2+1；
   2. 如果s1 > s2, 那么：如果m >= n1, m插入到最小堆，s2=s2+1；否则，n1pop出最大堆，然后push进最小堆，然后m进最大堆，s2=s2+1;
   3. 如果s1 < s2, 那么：如果m <=n2, m插入到最大堆，s1=s1+1；否则，n2pop出最小堆，然后push进最大堆，然后m进最小堆，s1=s1+1;

代码如下：

int main(int argc, char** argv) {
    priority_queue<int, vector<int>, less<int> > maxHeap;
    priority_queue<int, vector<int>, greater<int> > minHeap;
    
    vector<int> nums;
    srand(time(NULL));
    for (int i = 0; i < 10000; ++i) {
        nums.push_back(rand() % 1000);
        cout << nums[i] << " ";
    }
    cout << endl;
    
    double median; 
    for (int i = 0; i < nums.size(); ++i) {
        if (maxHeap.size() == minHeap.size()) {
            if (minHeap.empty() || nums[i] <= minHeap.top()) {
                maxHeap.push(nums[i]);
                median = maxHeap.top();
            } else {
                minHeap.push(nums[i]);
                median = minHeap.top();
            }
        } else if (maxHeap.size() > minHeap.size()) {
            if (maxHeap.empty() || nums[i] >= maxHeap.top()) {
                minHeap.push(nums[i]);
            } else {
                minHeap.push(maxHeap.top()); 
                maxHeap.pop();
                maxHeap.push(nums[i]);
            }
            median = (maxHeap.top() + minHeap.top()) / 2.0;
        } else {
            if (minHeap.empty() || nums[i] <= minHeap.top()) {
                maxHeap.push(nums[i]);
            } else {
                maxHeap.push(minHeap.top());
                minHeap.pop();
                minHeap.push(nums[i]);
            }
            median = (maxHeap.top() + minHeap.top()) / 2.0;
        }
        cout << i << "-th: " << median << endl;
    }
    
    return 0;
}