Sliding Window Maximum
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原题:
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?
解题:
用的暴力方法,O(N2)的时间复杂度,不过也AC了,暂时没有找到线性的O(N)的方法。
vector<int> maxSlidingWindow(vector<int>& nums, int k) { vector<int> ret; int length = nums.size(); if(length < 0 || k < 1 || k > length) return ret; for(int i=0; i<length-k+1; i++){ int max = nums[i]; for(int j=i+1; j<k+i; j++){ if(max < nums[j]) max = nums[j]; } ret.push_back(max); } return ret; }
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