LeetCode-in-Cpp
239. Sliding Window Maximum
Hard
You are given an array of integers 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.
Example 1:
Input: nums = [1,3,-1,-3,5,3,6,7], 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
Example 2:
Input: nums = [1], k = 1
Output: [1]
Example 3:
Input: nums = [1,-1], k = 1
Output: [1,-1]
Example 4:
Input: nums = [9,11], k = 2
Output: [11]
Example 5:
Input: nums = [4,-2], k = 2
Output: [4]
Constraints:
1 <= nums.length <= 105-104 <= nums[i] <= 1041 <= k <= nums.length
Solution
#include <vector>
#include <deque>
class Solution {
public:
std::vector<int> maxSlidingWindow(std::vector<int>& nums, int k) {
int n = nums.size();
std::vector<int> res(n - k + 1);
int x = 0;
std::deque<int> dq;
int i = 0;
int j = 0;
while (j < nums.size()) {
while (!dq.empty() && dq.back() < nums[j]) {
dq.pop_back();
}
dq.push_back(nums[j]);
if (j - i + 1 == k) {
res[x] = dq.front();
++x;
if (dq.front() == nums[i]) {
dq.pop_front();
}
++i;
}
++j;
}
return res;
}
};