Problem Description
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 ≤ inputarray’s size for non-empty array.
Follow up: Could you solve it in linear time?
Solution
- C++
vector<int> maxSlidingWindow(vector<int> & nums, int k) {
// A container for index in the current window
deque<int> dq;
// A container for max values as a result
vector<int> result;
// Iterate from begin to end in the array
for (int i = 0; i<nums.size(); i++) {
// Erase value in the front as the window move forward
if (!dq.empty() && dq.front() == i-k) dq.pop_front();
// Keep only the index of the largest number in the window and
// the index of the rightmost number in the window in container dq
while (!dq.empty() && nums[dq.back()] < nums[i]) dq.pop_back();
dq.push_back(i);
// Push the max value to the result vector
if (i >= k-1) result.push_back(nums[dq.front()]);
}
return result;
}