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. Return the max sliding window.

Example:

Input: nums = [1,3,-1,-3,5,3,6,7], and 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

Note:
You may assume k is always valid, 1 ≤ k ≤ input array's size for non-empty array.

Follow up:
Could you solve it in linear time?

思路

1. 维护一个双端队列，使队列中的值保持递减，即队首为最大值；
2. 如果遍历到的值的index >=k-1，即已经滑动到一个窗口的大小，则把队首元素加入结果；
3. 如果此时队列长度已经大于k，则把队首元素移除。

    def maxSlidingWindow2(self,nums,k):
        indices = deque()
        result = []
        for i in range(len(nums)):
            while indices and nums[i]>nums[indices[-1]]:
                indices.pop()
            indices.append(i)
            if i >= k-1:
                result.append(nums[indices[0]])
            if i-k+1 == indices[0]:
                indices.popleft()
        return result

另一种则为暴力法，直接遍历找最大。

    def maxSlidingWindow(self, nums, k):
        if not nums:
            return nums
        result = []
        for i in range(len(nums)-k+1):
            if i + k <= len(nums):
                subList = nums[i:i+k]
                result.append(max(subList))
            else:
                subList = nums[i:]
                result.append(max(subList))
        return result