forked from nyuhuyang/LeetCode
-
Notifications
You must be signed in to change notification settings - Fork 0
/
sliding-window-maximum.py
49 lines (44 loc) · 1.37 KB
/
sliding-window-maximum.py
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
# Time: O(n)
# Space: O(k)
# 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?
from collections import deque
class Solution(object):
def maxSlidingWindow(self, nums, k):
"""
:type nums: List[int]
:type k: int
:rtype: List[int]
"""
dq = deque()
max_numbers = []
for i in xrange(len(nums)):
while dq and nums[i] >= nums[dq[-1]]:
dq.pop()
dq.append(i)
if i >= k and dq and dq[0] <= i - k:
dq.popleft()
if i >= k - 1:
max_numbers.append(nums[dq[0]])
return max_numbers