标签：odi complex other ons you class time and ==
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.
1 class Solution: 2 def minSubArrayLen(self, mins, nums): 3 """ 4 :type s: int 5 :type nums: List[int] 6 :rtype: int 7 """ 8 if not nums: return 0 9 l, r, s, m = 0, 0, 0, None 10 while r < len(nums): 11 s += nums[r] 12 r += 1 13 while s >= mins: 14 m = r - l if not m else min(m, r-l) 15 s -= nums[l] 16 l += 1 17 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.
1 # myself solution, but can not pass all the test cases 2 3 if not nums: return 0 4 l, r, ans = 0, len(nums)-1, 0 5 while l<= r and sum(nums[l:r+1]) >= s: 6 ans = r-l +1 7 #print(nums[l:r+1]) 8 if nums[l] < nums[r]: 9 l += 1 10 elif nums[l] > nums[r]: 11 r -= 1 12 else: 13 templ, tempr = l+1, r-1 14 condition = True 15 while(templ <= tempr and condition): 16 if nums[templ] == nums[tempr]: 17 templ += 1 18 tempr -= 1 19 elif nums[templ] < nums[tempr]: 20 l += 1 21 condition = False 22 else: 23 r -= 1 24 condition = False 25 if condition: 26 l += 1 27 return ans
[LeetCode] 209. Minimum Size Subarray Sum_Medium
标签：odi complex other ons you class time and ==
原文地址：https://www.cnblogs.com/Johnsonxiong/p/9189709.html
