[leetcode ]239. Sliding Window Maximum

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
---------------               -----
[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?

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

Solution:

vector<int> maxSlidingWindow(vector<int>& nums, int k)
{
deque<int> idq;
vector<int> result;
int i = 0;

for (i = 0; i < k - 1; i++)
{
while (!idq.empty() && (idq.back() < nums[i]))
idq.pop_back();
idq.push_back(nums[i]);
}

for (i = k - 1; i < nums.size(); i++)
{
while (!idq.empty() && (idq.back() < nums[i]))
idq.pop_back();
idq.push_back(nums[i]);

result.push_back(idq.front());

if (idq.front() == nums[i - k + 1])
idq.pop_front();
}

return result;
}