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Copy path239 Sliding Window Maximum.py
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239 Sliding Window Maximum.py
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# -*- coding: utf-8 -*-
"""
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?
Author: Rajeev Ranjan
"""
class Solution:
def maxSlidingWindow(self, nums, k):
"""
Algorithms
1. brute force
2. heap with lazy deletion
3. queue
In the double-ended queue algorithm, you need to assign the queue with a meaning, in a way that you can
access the window's max in O(1).
Invariant: the queue is storing non-decreasing-ordered elements of current window.
The queue stores the index rather than element
:type nums: list[]
:type k: int
:rtype: list[]
"""
q = [] # store the index
ret = []
n = len(nums)
for i in xrange(n):
while q and q[0] <= i-k:
q.pop(0)
while q and nums[q[-1]] < nums[i]:
q.pop()
q.append(i)
if i >= k-1:
ret.append(nums[q[0]])
return ret