[LeetCode] 209. Minimum Size Subarray Sum_Medium

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