239. Sliding Window Maximum

[huimeich]

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

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[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?

[huimeich]

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.

[huimeich]

class Solution {
public:
    vector<int> maxSlidingWindow(vector<int>& nums, int k) {
        deque<int> q;
        vector<int> res;
        for (int i = 0; i < nums.size(); i++) {
            if (i >= k && q.front() == nums[i - k]) {
                q.pop_front();
            }
            if (q.empty() || q.front() < nums[i]) {
                q.clear();
                q.push_back(nums[i]);
            } else {
                for (int j = 0; j < q.size(); j++) {
                    while (j < q.size() && q[j] < nums[i]) {
                        q.pop_back();
                    }
                }
                q.push_back(nums[i]);
            }
            if (i >= k-1) res.push_back(q.front());
        }
        return res;
    }
};