LeetCode 239. Sliding Window Maximum

Posted on Edited on In 
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
class Solution {
public:
    vector<int> maxSlidingWindow(vector<int>& nums, int k) {
        priority_queue<pair<int, int>> q;
        for (int i = 0; i < k; ++i)
            q.emplace(nums[i], i);
        vector<int> ans = {q.top().first};
        for (int i = k; i < nums.size(); ++i) {
            q.emplace(nums[i], i);
            while (q.top().second <= i - k)
                q.pop();
            ans.push_back(q.top().first);
        }
        return ans;
    }
};
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
class Solution {
public:
    vector<int> maxSlidingWindow(vector<int>& nums, int k) {
        deque<int> q;
        for (int i = 0; i < k; ++i) {
            while (!q.empty() && nums[i] >= nums[q.back()])
                q.pop_back();
            q.push_back(i);
        }
        vector<int> ans{nums[q.front()]};
        for (int i = k; i < nums.size(); ++i) {
            while (!q.empty() && nums[i] >= nums[q.back()])
                q.pop_back();
            q.push_back(i);
            while (q.front() <= i - k)
                q.pop_front();
            ans.push_back(nums[q.front()]);
        }
        return ans;
    }
};
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
class Solution {
public:
    vector<int> maxSlidingWindow(vector<int>& nums, int k) {
        const int n = nums.size();
        vector<int> prefixMax(n), suffixMax(n);
        for (int i = 0; i < n; ++i)
            if (i % k == 0) prefixMax[i] = nums[i];
            else prefixMax[i] = max(prefixMax[i - 1], nums[i]);
        suffixMax[n - 1] = nums[n - 1];
        for (int i = n - 2; i >= 0; --i)
            if ((i + 1) % k == 0) suffixMax[i] = nums[i];
            else suffixMax[i] = max(suffixMax[i + 1], nums[i]);
        vector<int> ans;
        for (int i = 0; i <= n - k; ++i)
            ans.push_back(max(suffixMax[i], prefixMax[i + k - 1]));
        return ans;
    }
};