leetcode 239. Sliding Window Maximum

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 the k numbers in the window. Each time the sliding window moves right by one position.


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


Window position                Max
---------------               -----
[1  3  -1] -3  5  3  6  7       3
 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       5
 1  3  -1  -3 [5  3  6] 7       6
 1  3  -1  -3  5 [3  6  7]      7
Therefore, return the max sliding window as [3,3,5,5,6,7].


Note: 
You may assume k is always valid, ie: 1 ≤ k ≤ input array's size for non-empty array.


Follow up:
Could you solve it in linear time?

#include "stdafx.h"
#include<vector>
#include<set>
#include<map>
#include <functional> 
#include<iostream>
using namespace std;
class Solution {
public:
	vector<int> maxSlidingWindow(vector<int>& nums, int k) {
		vector<int>re;
		if(nums.empty())
			return re;
		if(k==1)
			return nums;
		set<int,std::greater<int>>aa;
		map<int,int>count;
		for(int i=0;i<k;i++)
		{
			count[nums[i]]++;
			aa.insert(nums[i]);
		}
		re.push_back(*aa.begin());
		for(int i=1;i<=nums.size()-k;i++)
		{
			count[nums[i-1]]--;
			if(count[nums[i-1]]==0)
				aa.erase(aa.find(nums[i-1]));
			count[nums[i+k-1]]++;
			aa.insert(nums[i+k-1]);
			re.push_back(*aa.begin());
		}
		return re;
	}
};

int _tmain(int argc, _TCHAR* argv[])
{
	Solution sl;
	int aa[6]={1,3,1,2,0,5};
	vector<int>nums(aa,aa+6);
	vector<int>re=sl.maxSlidingWindow(nums,3);

	system("pause");
	return 0;
}

accepted