209. Minimum Size Subarray Sum刷题笔记

时间复杂度为n是因为尽管套了两个循环，但是每个元素最多只是进来被操作一次，出去被操作一次

class Solution:
    def minSubArrayLen(self, target: int, nums: List[int]) -> int:
        result = 100001
        sum_num = 0
        j = -1
        for i in range(len(nums)):
            sum_num += nums[i]
            while sum_num>=target:
                j += 1
                sum_num -=  nums[j]                
                result = min(result,i-j+1)
        if result == 100001:
            return 0
        return result