#Leetcode# 209. Minimum Size Subarray Sum
https://leetcode.com/problems/minimum-size-subarray-sum/
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).
$O(n)$ 代码:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
class Solution { public: int minSubArrayLen(int s, vector<int>& nums) { if(nums.empty()) return 0; int n = nums.size(); int cnt = n + 1; int R = 0, L = 0, sum = 0; while(R < n) { while(sum < s && R < n) sum += nums[R ++]; while(sum >= s) { cnt = min(cnt, R - L); sum -= nums[L ++]; } } if(cnt == n + 1) return 0; return cnt; } };
$O(n^2)$ 代码:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
class Solution { public: int minSubArrayLen(int s, vector<int>& nums) { if(nums.empty()) return 0; int n = nums.size(); int cnt = n + 1; for(int i = 0; i < n; i ++) { int temp = i; int ans = 0; while(ans < s && temp < n) { ans += nums[temp]; temp ++; } if(ans < s) continue; else cnt = min(cnt, temp - i); } if(cnt == n + 1) return 0; return cnt; } };