#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：

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：

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;
    }
};