[leetcode][deque] Sliding Window Maximum

题目：

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?

Hint:

1. How about using a data structure such as deque (double-ended queue)?
2. The queue size need not be the same as the window’s size.
3. Remove redundant elements and the queue should store only elements that need to be considered.

class Solution {public:vector<int> maxSlidingWindow(vector<int>& nums, int k) {vector<int> res;int n = nums.size();if (nums.empty() || k <= 0 || k > n) return res;deque<int> deq;//deq表示元素在nums中的位置deq.push_back(0);for (int i = 1; i < k; ++i){while (!deq.empty() && nums[i] >= nums[deq.back()]) deq.pop_back();//去除队列中所有比当前元素小的元素（这些元素比当前元素小又比当前元素早出现，当前元素存在期间它们绝不可能成为max）deq.push_back(i);}res.push_back(nums[deq.front()]);//队首元素是窗口中的max（否则会被后面的元素删除的）for (int i = k; i < n; ++i){if (i - deq.front() >= k) deq.pop_front();//如果队首元素已不在窗口中，将其删除while (!deq.empty() && nums[i] >= nums[deq.back()]) deq.pop_back();//删除队列中所有比当前元素小的元素deq.push_back(i);//当前元素入队res.push_back(nums[deq.front()]);//输出这个窗口的max}return res;}};

注：线性时间复杂度