239. Sliding Window Maximum
Posted ZHOU YANG
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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:
- How about using a data structure such as deque (double-ended queue)?
- The queue size need not be the same as the window’s size.
- Remove redundant elements and the queue should store only elements that need to be considered.
思路:按照提示,利用deque储存数组的下标,保持deque的头部始终存储当前窗口最大的数的下标。每次循环,剔除deque中比nums[i]小的数。将窗口右边界数的下标加入deque中,同时剔除deque头部越界的序号。当遍历的数已经达到窗口的大小时,此时deque头部的数的下标就是当前窗口的最大的那个数的下标,将这个数加入最终返回的那个数组中。
class Solution { public: vector<int> maxSlidingWindow(vector<int>& nums, int k) { vector<int> ret; deque<int>q; for(int i=0;i<nums.size();i++){ //只在deque中保留最大的数 while(!q.empty()&&nums[i]>=nums[q.back()]) q.pop_back(); //每次后移窗口,加入i q.push_back(i); //将越界的front剔除 if(q.front()<=i-k) q.pop_front(); //当达到窗口大小的时候,将deque中的最大值加入ret数组 if(i>=k-1) ret.push_back(nums[q.front()]); } return ret; } };
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