【剑指offer】面试题41：数据流中的中位数——大、小堆实现

方法一：排序取中位数（超时）

class MedianFinder {
    
public:
    /** initialize your data structure here. */
    vector<int> record;
    MedianFinder() {
    

    }
    
    void addNum(int num) {
    
        record.push_back(num);
    }
    
    double findMedian() {
    
        double result = 0;
        if( record.size() == 0 )
            return result;
        sort( record.begin(), record.end() );
        //奇数
        if( record.size() & 1 )
        {
    
            result = record[ record.size() / 2 ];
        }
        else
        {
    
            result = (double)( record[ record.size() / 2 - 1 ] + record[ record.size() / 2 ] ) / 2;
        }

        return result;
    }
};

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

方法二：大、小堆实现

1.通过两个堆，一个最大堆一个最小堆，大小不超过1 ，如果超过1就把大小堆平衡一下
2.中位数如果是奇数则为最大堆的top或最小堆的top，如果最大堆和最小堆的大小相等，则取两个堆顶的平均值
3.最小堆里面的数据都要比最大堆大

class MedianFinder {
    
public:
    /** initialize your data structure here. */

    //最大堆
    priority_queue< int, vector<int> > max_heap;
    //最小堆
    priority_queue< int, vector<int>, greater<int> > min_heap;
    MedianFinder() {
    

    }
    
    void addNum(int num) {
    

        
        if( max_heap.empty() || num < max_heap.top() )
        {
    
            max_heap.push( num );
        }
        else
            min_heap.push( num );

        //维持堆得大小
        if( max_heap.size() == min_heap.size() + 2)
        {
    
            min_heap.push(max_heap.top());
            max_heap.pop();
        }
        else if( min_heap.size() == max_heap.size() + 2)
        {
    
            max_heap.push( min_heap.top() );
            min_heap.pop();
        }

    }
    
    double findMedian() {
    

        if( ( max_heap.size() + min_heap.size() ) == 0 )
            return -1;
        if( max_heap.size() > min_heap.size() )
        {
    
            return max_heap.top();
        }
        else if( max_heap.size() < min_heap.size() )
        {
    
            return min_heap.top();
        }
        else
            return double( max_heap.top() + min_heap.top() ) / 2;

    }
};