Median is the middle value in an ordered integer list. If the size of the list is even, there is no middle value. So the median is the mean of the two middle value.

[2,3,4] , the median is 3

[2,3], the median is (2 + 3) / 2 = 2.5

Given an array nums, there is a sliding window of size k which is moving from the very left of the array to the very right. You can only see thek numbers in the window. Each time the sliding window moves right by one position. Your job is to output the median array for each window in the original array.

For example,Given nums = [1,3,-1,-3,5,3,6,7], and k = 3.

Window position                Median---------------               -----[1  3  -1] -3  5  3  6  7       1 1 [3  -1  -3] 5  3  6  7       -1 1  3 [-1  -3  5] 3  6  7       -1 1  3  -1 [-3  5  3] 6  7       3 1  3  -1  -3 [5  3  6] 7       5 1  3  -1  -3  5 [3  6  7]      6
Therefore, return the median sliding window as [1,-1,-1,3,5,6].
Note: You may assume k is always valid, ie: 1 ≤ k ≤ input array's size for non-empty array.
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給出一個序列nums和一個窗口值k，每k個數(shù)求其中的中位數(shù)。前天剛做了類似的題（295題），求一個動態(tài)增加元素的序列的中位數(shù)，這里能用295題的想法，然后還要添加刪除元素的功能。
代碼：
class Solution{public:	vector<double> medianSlidingWindow(vector<int>& nums, int k)	{		int i;		vector<double> res;		for(i = 0; i < k; ++i)		{			addNum(nums[i]);		}		for(i; i < nums.size(); ++i)		{			res.push_back(findMedian());			delNum(nums[i-k]);			addNum(nums[i]);		}		res.push_back(findMedian());		return res;	}	void addNum(int num)	{		double n = double(num);		if(set1.empty()) set1.insert(n);		else if(set1.size() > set2.size())		{			if(n >= *set1.rbegin())			{				set2.insert(n);			}			else			{				set2.insert(*set1.rbegin());				set1.erase(PRev(set1.end()));				set1.insert(n);			}		}		else		{			if(n <= *set2.begin())			{				set1.insert(n);			}			else			{				set1.insert(*set2.begin());				set2.erase(set2.begin());				set2.insert(n);			}		}	}	void delNum(int num)	{		double n = double(num);		auto iter1 = set1.find(n), iter2 = set2.find(n);		if(iter1 != set1.end())		{			set1.erase(iter1);			if(set1.size() < set2.size())			{				set1.insert(*set2.begin());				set2.erase(set2.begin());			}		}		else		{			set2.erase(iter2);			if(set1.size() > set2.size() + 1)			{				set2.insert(*set1.rbegin());				set1.erase(prev(set1.end()));			}		}	}	double findMedian()	{		if(set1.size() > set2.size())		{			return *set1.rbegin();		}		else		{			return (*set1.rbegin() + *set2.begin()) / 2;		}	}private:	multiset<double> set1, set2;};