class Solution:
def longestSubarray(self, nums: List[int], limit: int) -> int:
#利用第三方库,将列表有序化,底层其实是一个最小堆
from sortedcontainers import SortedList
left, right = 0, 0
size = len(nums)
s = SortedList()
res = 1
while right<size:
s.add(nums[right])
while s[-1]-s[0]>limit:
s.remove(nums[left])
left+=1
res = max(res,right-left+1)
right+=1
return res
#超出时间限制
res = 1
left,right = 0,0
size = len(nums)
win = deque()
Max,Min = float('-inf'),float('inf')
while right<size:
win.append(nums[right])
Max = max(Max,nums[right])
Min = min(Min,nums[right])
while Max-Min>limit:
win.popleft()
Max = max(win)
Min = min(win)
res = max(res,len(win))
right+=1
return res
#官方解答,还可以利用两个单调队列去维护最大值和最小值
n = len(nums)
queMax, queMin = deque(), deque()
left = right = ret = 0
while right < n:
while queMax and queMax[-1] < nums[right]:
queMax.pop()
while queMin and queMin[-1] > nums[right]:
queMin.pop()
queMax.append(nums[right])
queMin.append(nums[right])
while queMax and queMin and queMax[0] - queMin[0] > limit:
if nums[left] == queMin[0]:
queMin.popleft()
if nums[left] == queMax[0]:
queMax.popleft()
left += 1
ret = max(ret, right - left + 1)
right += 1
return ret
- 题目的本质就是需要维护一个窗口中最大值和最小值,最大值和最小值的差不超过limit即可
- 利用python第三方库去解决这个问题,不管每次添加或者删除元素之和这个列表仍然是有序的
- 这个库还有三个
-
from sortedcontainers import SortedList
-
from sortedcontainers import SortedDict
-
from sortedcontainers import SortedSet