[leetcode] 209. Minimum Size Subarray Sum
阿新 • • 發佈:2022-03-20
題目
Given an array of positive integers nums
and a positive integer target
, return the minimal length of a contiguous subarray [numsl, numsl+1, ..., numsr-1, numsr]
of which the sum is greater than or equal to target
. If there is no such subarray, return 0
instead.
Example 1:
Input: target = 7, nums = [2,3,1,2,4,3] Output: 2 Explanation: The subarray [4,3] has the minimal length under the problem constraint.
Example 2:
Input: target = 4, nums = [1,4,4]
Output: 1
Example 3:
Input: target = 11, nums = [1,1,1,1,1,1,1,1]
Output: 0
Constraints:
1 <= target <= 109
1 <= nums.length <= 105
1 <= nums[i] <= 105
Follow up: If you have figured out the O(n)
solution, try coding another solution of which the time complexity is O(n log(n))
思路
滑動視窗,當視窗內數字和大於等於target時,滑動左視窗至數字和小於target,將此時視窗大小記錄下來,等到迴圈結束,所有視窗大小的最小值就是結果。
程式碼
python版本:
class Solution: def minSubArrayLen(self, target: int, nums: List[int]) -> int: cnt = 0 minl = math.inf l = 0 for r in range(len(nums)): cnt += nums[r] if cnt >= target: while cnt >= target and l <= r: cnt -= nums[l] l += 1 minl = min(minl, r-l+2) return 0 if minl == math.inf else minl