[leetcode] 209. Minimum Size Subarray Sum

题目

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

滑动窗口，当窗口内数字和大于等于target时，滑动左窗口至数字和小于target，将此时窗口大小记录下来，等到循环结束，所有窗口大小的最小值就是结果。

代码

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

​