LeetCode:209. Minimum Size Subarray Sum

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题目描述

Given an array of n positive integers and a positive integer s, find the minimal length of a contiguous subarray of which the sum ≥ s. If there isn’t one, return 0 instead.

Example: 

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Input: s = 7, nums = [2,3,1,2,4,3]
Output: 2
Explanation: the subarray [4,3] has the minimal length under the problem constraint.

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).

题目要求在一串数字中找到最短的连续数字，让他们的和大于给定的正数。

这里需要记录数组的开始和结尾的索引，并且需要记录连续的数字之和。每当数字比给定目标小的时候，右边的索引要往右移动，其他情况则左边索引向右移动一格，并且如果比最小的结果更小的时候需要记录这个长度。

代码实现

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class Solution:
    def minSubArrayLen(self, s: int, nums: 'list[int]') -> int:
        if len(nums) == 0:
            return 0
        
        start = 0
        end = 1
        sum = nums[0]
        result = len(nums)
        find = False
        while True:
            if sum >= s:
                find = True
                result = end - start if end - start < result else result
                if start < len(nums)-1:
                    sum -= nums[start]
                    start += 1
                else:
                    break
            else:
                if end < len(nums):
                    sum += nums[end]
                    end += 1
                else:
                    break

        return 0 if find == False else result