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: 2Explanation: 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：

class Solution {public:    int minSubArrayLen(int s, vector
      
       & 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：

class Solution {public:    int minSubArrayLen(int s, vector
      
       & 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;    }};