leetcode 239. Sliding Window Maximum 双端队列 滑动窗口最大值

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?

思路： 这道题目的讲解看代码面试指南。维护一个双端队列，这个队列保存依次遍历的下标值。入队列的规则是比队尾小的数可以入队，比队尾大的数，将队列末尾所有小的数都清除直到遇到比这个数大的数位置。每到一个窗口，都从队头读取元素，如果元素已经超出窗口范围，就在看下一个元素。

class Solution {
public:
vector<int> maxSlidingWindow(vector<int>& nums, int k) {
deque<int> q;
int n=nums.size();
vector<int> res;
if(n==0||k<1||n<k) return res;
for(int i=0;i<n;i++)
{
while(!q.empty()&&nums[q.back()]<=nums[i])
q.pop_back();
q.push_back(i);
if(q.front()==i-k)
q.pop_front();
if(i>=k-1)
res.push_back(nums[q.front()]);
}
return res;
}
};