力扣 面試題41. 數據流中的中位數 優先隊列+思維

https://leetcode-cn.com/problems/shu-ju-liu-zhong-de-zhong-wei-shu-lcof/
思路：搞兩個堆，假設當前一共有$n$個元素，大根堆存儲最小的$(n+1)/2$個元素，小根堆存儲最大的$n/2$個元素，每次插入元素的時候維護這兩個堆就行了，可以證明插入後修改的複雜度爲$O(lgn)$，得到中位數的複雜度爲$O(1)$。

class MedianFinder {
public:
    /** initialize your data structure here. */
    priority_queue<int> big;
    priority_queue<int,vector<int>,greater<int>> small;
    int siz;
    MedianFinder() {
        siz=0;
    }
    
    void addNum(int num) {
        ++siz;
        big.push(num);
        int res=(siz+1)>>1;
        while(big.size()>res)
            small.push(big.top()),big.pop();
        while(!small.empty()&&big.top()>small.top())
        {
            int val1=big.top(),val2=small.top();
            big.pop(),small.pop();
            big.push(val2);
            small.push(val1);
        } 
    }
    
    double findMedian() {
        if(siz==0)
            return 0;
        else if(siz&1)
            return big.top();
        else
            return (big.top()+small.top())*1.0/2;
    }
};

/**
 * Your MedianFinder object will be instantiated and called as such:
 * MedianFinder* obj = new MedianFinder();
 * obj->addNum(num);
 * double param_2 = obj->findMedian();
 */