# Sliding Window Maximum Problem

## Description

LeetCode Problem 239.

You are given an array of integersnums, 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:

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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:

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Input: nums = [1], k = 1
Output: [1]
``````

Example 3:

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Input: nums = [1,-1], k = 1
Output: [1,-1]
``````

Example 4:

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Input: nums = [9,11], k = 2
Output: [11]
``````

Example 5:

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Input: nums = [4,-2], k = 2
Output: [4]
``````

Constraints:

• 1 <= nums.length <= 10^5
• -10^4 <= nums[i] <= 10^4
• 1 <= k <= nums.length

## Sample C++ Code

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class Solution {
public:
vector<int> maxSlidingWindow(vector<int>& nums, int k) {
int len = nums.size();
vector<int> ans;
deque<int> maxs;

for (int i = 0; i < len; i ++) {
while (!maxs.empty() && maxs.back() < nums[i]) maxs.pop_back();
maxs.push_back(nums[i]);

if (i >= k && nums[i-k] == maxs.front())
maxs.pop_front();
if (i >= k - 1)
ans.push_back(maxs.front());
}

return ans;
}
};
``````