思路:维护一个长度为k的单调双端队列,队首元素为最大值的下标
class Solution {
public:
vector<int> maxInWindows(vector<int>& nums, int k) {
vector<int> res; //res存滑动窗口最大值
deque<int> q; //q存下标
for (int i = 0; i < nums.size(); i ++ )
{
if (q.size() && q.front() <= i - k) q.pop_front(); //滑出窗口
while (q.size() && nums[q.back()] <= nums[i]) q.pop_back(); //维护队列单调性
q.push_back(i);
if (i >= k - 1) res.push_back(nums[q.front()]); //最大值请入res
}
return res;
}
};