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:

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). 
 
 class Solution:
def minSubArrayLen(self, mins, nums):
"""
:type s: int
:type nums: List[int]
:rtype: int
"""
if not nums: return 0
l, r, s, m = 0, 0, 0, None
while r < len(nums):
s += nums[r]
r += 1
while s >= mins:
m = r - l if not m else min(m, r-l)
s -= nums[l]
l += 1
return m if m else 0

我自己最开始的solution想的是用Two Pointers, l = 0, r = len(nums) -1, 然后分别往中间扫, 但是pass不了所有的test cases, 不知道问题出在哪.

         # myself solution, but can not pass all the test cases

         if not nums: return 0
l, r, ans = 0, len(nums)-1, 0
while l<= r and sum(nums[l:r+1]) >= s:
ans = r-l +1
#print(nums[l:r+1])
if nums[l] < nums[r]:
l += 1
elif nums[l] > nums[r]:
r -= 1
else:
templ, tempr = l+1, r-1
condition = True
while(templ <= tempr and condition):
if nums[templ] == nums[tempr]:
templ += 1
tempr -= 1
elif nums[templ] < nums[tempr]:
l += 1
condition = False
else:
r -= 1
condition = False
if condition:
l += 1
return ans

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