[email protected] [239]Sliding Window Maximum

https://leetcode.com/problems/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 knumbers 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?

Hint: How about using a data structure such as deque (double-ended queue)?

class Solution {
public:
    vector<int> maxSlidingWindow(vector<int>& nums, int k) {
        vector<int> ret; ret.clear();
        if(nums.size() == 0) return ret;

        multiset<int> myset; myset.clear();

        for(int i=0;i<=k-1;++i) myset.insert(nums[i]);
        multiset<int>::reverse_iterator pt = myset.rbegin();
        ret.push_back(*pt);

        for(int i=k;i<nums.size();++i){
            myset.erase(myset.find(nums[i-k]));
            myset.insert(nums[i]);
            pt = myset.rbegin();
            ret.push_back(*pt);
        }

        return ret;
    }
};