動機
sliding window但是用heap去找window的最大最小
Problem
Given an array of integers nums
and an integer limit
, return the size of the longest non-empty subarray such that the absolute difference between any two elements of this subarray is less than or equal to limit
.
Example 1:
Input: nums = [8,2,4,7], limit = 4Output: 2 Explanation: All subarrays are: [8] with maximum absolute diff |8-8| = 0 <= 4.[8,2] with maximum absolute diff |8-2| = 6 > 4. [8,2,4] with maximum absolute diff |8-2| = 6 > 4.[8,2,4,7] with maximum absolute diff |8-2| = 6 > 4.[2] with maximum absolute diff |2-2| = 0 <= 4.[2,4] with maximum absolute diff |2-4| = 2 <= 4.[2,4,7] with maximum absolute diff |2-7| = 5 > 4.[4] with maximum absolute diff |4-4| = 0 <= 4.[4,7] with maximum absolute diff |4-7| = 3 <= 4.[7] with maximum absolute diff |7-7| = 0 <= 4. Therefore, the size of the longest subarray is 2.
Example 2:
Input: nums = [10,1,2,4,7,2], limit = 5Output: 4 Explanation: The subarray [2,4,7,2] is the longest since the maximum absolute diff is |2-7| = 5 <= 5.
Example 3:
Input: nums = [4,2,2,2,4,4,2,2], limit = 0Output: 3
Constraints:
1 <= nums.length <= 105
1 <= nums[i] <= 109
0 <= limit <= 109
Sol
from sortedcontainers import SortedList
class Solution:
def longestSubarray(self, nums: List[int], limit: int) -> int:
win = SortedList()
i, ret = 0, 0
for j,n in enumerate(nums):
win.add(n)
while i<=j and win[-1]-win[0] > limit:
win.remove(nums[i])
i += 1
ret = max(ret, j-i+1)
return ret