leetcode - Sliding Window Maximum
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题目:
Sliding Window Maximum
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, 1 ≤ k ≤ input array's size.
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.
分析:
用一个双向队列d保存窗口信息,队列中的元素是非递增的(即相邻严肃要么相等要么递减),队列的元素个数不超过窗口大小,队列的d.front()是当前窗口的最大值。
class Solution {public: vector<int> maxSlidingWindow(vector<int>& nums, int k) { if(k<1 || nums.size()<k) return vector<int>(); deque<int> d; vector<int> res; for(int i=0;i<k;++i) { while(!d.empty() && d.back()<nums[i]) d.pop_back(); d.push_back(nums[i]); } res.push_back(d.front()); for(int i=k;i<nums.size();++i) { if(nums[i-k]==d.front()) d.pop_front(); while(!d.empty() && d.back()<nums[i]) d.pop_back(); d.push_back(nums[i]); res.push_back(d.front()); } return res; }};
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