Problem:
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. Return the max sliding window.
Follow up:
Could you solve it in linear time?
Example:
Input: nums = [1,3,-1,-3,5,3,6,7], and k = 3
Output: [3,3,5,5,6,7]
Explanation:
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
Constraints:
1 <= nums.length <= 10^5-10^4 <= nums[i] <= 10^41 <= k <= nums.length
思路：
Solution (C++):
vector<int> maxSlidingWindow(vector<int>& nums, int k) {
deque<int> q;
int n = nums.size();
vector<int> res(n-k+1, 0);
for (int i = 0; i < n; ++i) {
while (!q.empty() && nums[i] >= nums[q.back()]) {
q.pop_back();
}
q.push_back(i);
if (q.front() == i-k) q.pop_front();
if (i >= k-1) res[i-k+1] = nums[q.front()];
}
return res;
}
性能：
Runtime: 60 ms Memory Usage: 10.5 MB
思路：
Solution (C++):
性能：
Runtime: ms Memory Usage: MB
思路：
Solution (C++):
性能：
Runtime: ms Memory Usage: MB
