> For the complete documentation index, see [llms.txt](https://kongchengzhuge.gitbook.io/leetcode/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://kongchengzhuge.gitbook.io/leetcode/209.-minimum-size-subarray-sum.md).

# 209. Minimum Size Subarray Sum

## problem description

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).
```

## algorithm thought

这里求出连续子数组之和大于s的最小值。有两种方法，题目中follow up也提出，有O(n)和O(nlgn)的办法。

首先说下O(nlgn)，这种形式的时间复杂度一般是分治法，类似的有归并排序。对于这题，连续数组要不就是包含中间那个值，或者是在左半数组中或者是在右半数组中。所以首先处理连续数组在中间的情况，从中间开始左右扩展，然后左边右边递归处理即可。最后得到最大的

然后是O(n)的解法，这里还是使用快慢指针，快的先加，如果超过了s，慢指针再往前，慢指针只需要减，直到小于s。直到快指针超过数组长度即可结束，中间每次比较得到最小值

## code

```cpp
class Solution {
public:
    int minSubArrayLen(int s, vector<int>& nums) {
        if(nums.size()==0)
            return 0;
        int fast=0,slow=0;
        int sum=0;
        int res=INT_MAX;
        while(fast<nums.size()){
            sum+=nums[fast];
            while(sum>=s){
                res=min(res,fast-slow+1);
                sum-=nums[slow++];    
            }
            fast++;
        }
        return res==INT_MAX?0:res;
    }
};
```

## algorithm analysis

thought中已经分析了时间复杂度，这里就不多说了