p146 在数据流中查找中位数(leetcode 295)

一：解题思路

方法一：每次添加一个数字，都使它成为一个递增有序的序列。时间复杂度分别为：addNum:O(n),findMedian:O(1)

方法二:采用一个最小堆和一个最大堆。时间复杂度分别为：addNum:O(log(n)),findMedian:O(1)

二：完整代码示例 (C++版和Java版)

方法一C++:

方法一Java:

class MedianFinder
    {
        private List<Integer> m_data=new ArrayList<>();

        /** initialize your data structure here. */
        public MedianFinder() {

        }

        public void addNum(int num)
        {
              int idx=m_data.size()-1;
              while (idx>=0 && m_data.get(idx)<num) idx--;
              m_data.add(idx+1,num);
        }

        public double findMedian()
        {
               int n=m_data.size();
               if((n&1)==1)
                   return m_data.get(n/2);
               else
                   return (m_data.get(n/2-1)+m_data.get(n/2))/2.0;
        }
    }

方法二C++:

class MedianFinder 
{
private:
    priority_queue<int, vector<int>, greater<int>> minHeap;
    priority_queue<int, vector<int>, less<int>> maxHeap;
public:
    /** initialize your data structure here. */
    MedianFinder() {

    }

    void addNum(int num) 
    {
        minHeap.push(num);
        maxHeap.push(minHeap.top());
        minHeap.pop();

        if (maxHeap.size() - minHeap.size() > 1)
        {
            minHeap.push(maxHeap.top());
            maxHeap.pop();
        }
    }

    double findMedian() 
    {
        if (maxHeap.size() > minHeap.size())
        {
            return maxHeap.top();
        }
        else
        {
            return (maxHeap.top()+minHeap.top()) / 2.0;
        }
    }
};

方法二Java:

class MedianFinder
    {
        private Queue<Integer> minHeap=new PriorityQueue<>();
        private Queue<Integer> maxHeap=new PriorityQueue<>(Collections.reverseOrder());

        /** initialize your data structure here. */
        public MedianFinder() {

        }

        public void addNum(int num)
        {
               minHeap.add(num);
               maxHeap.add(minHeap.poll());
               if(maxHeap.size()-minHeap.size()>1)
               {
                   minHeap.add(maxHeap.poll());
               }
        }

        public double findMedian()
        {
               if(maxHeap.size()>minHeap.size())
               {
                   return maxHeap.peek();
               }
               else
               {
                   return (maxHeap.peek()+minHeap.peek())/2.0;
               }
        }
    }