Sliding Window Maximum

原题：
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?

解题：
用的暴力方法，O（N2）的时间复杂度，不过也AC了，暂时没有找到线性的O(N)的方法。

    vector<int> maxSlidingWindow(vector<int>& nums, int k) {
        vector<int> ret;
        int length = nums.size();
        if(length < 0 || k < 1 || k > length)
            return ret;

        for(int i=0; i1; i++){
            int max = nums[i];
            for(int j=i+1; jif(max < nums[j])
                   max = nums[j];
            }
            ret.push_back(max);
        }

        return ret;
    }