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?
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这题的思路比较容易想到,但是反应比较慢。实际上就三步:1.左边看看是不是要被删了 2. 右边要把比当前元素小等的弹出来,再把最新的放进栈 3. 双端队列最左边是最大的。用Python的list模拟双端口队列。
class Solution:
def maxSlidingWindow(self, nums, k: int):
queue,qs,l,res = [],0,len(nums),[]
for i in range(l):
if (i >= k and i-k == queue[qs]): #pop
qs += 1
ql = len(queue)
while (ql > qs and nums[queue[-1]] <= nums[i]): #push #bug2: ql >= qs
queue.pop()
ql -= 1
queue.append(i) #bug1: queue.append(nums[i])
if (i >= k-1):
res.append(nums[queue[qs]])
return res
再贴一个来自Stephan的写法:
def maxSlidingWindow(self, nums, k):
d = collections.deque()
out = []
for i, n in enumerate(nums):
while d and nums[d[-1]] < n:
d.pop()
d += i,
if d[0] == i - k:
d.popleft()
if i >= k - 1:
out += nums[d[0]],
return out