# 0209. 长度最小的子数组
### 题目地址(209. 长度最小的子数组)
### 题目描述
```
给定一个含有 n 个正整数的数组和一个正整数 s ，找出该数组中满足其和 ≥ s 的长度最小的 连续 子数组，并返回其长度。如果不存在符合条件的子数组，返回 0。
示例：
输入：s = 7, nums = [2,3,1,2,4,3]
输出：2
解释：子数组 [4,3] 是该条件下的长度最小的子数组。
进阶：
如果你已经完成了 O(n) 时间复杂度的解法, 请尝试 O(n log n) 时间复杂度的解法。
```
### 前置知识
* [滑动窗口](/leetcode-solution/thinkings/slide-window.md)
### 公司
* 阿里
* 腾讯
* 百度
* 字节
### 思路
用滑动窗口来记录序列， 每当滑动窗口中的 sum 超过 s， 就去更新最小值，并根据先进先出的原则更新滑动窗口，直至 sum 刚好小于 s
> 这道题目和 leetcode 3 号题目有点像，都可以用滑动窗口的思路来解决
### 关键点
* 滑动窗口简化操作(滑窗口适合用于求解这种要求`连续`的题目)
### 代码
* 语言支持：JS，C++，Python
Python Code：
```python
class Solution:
def minSubArrayLen(self, s: int, nums: List[int]) -> int:
l = total = 0
ans = len(nums) + 1
for r in range(len(nums)):
total += nums[r]
while total >= s:
ans = min(ans, r - l + 1)
total -= nums[l]
l += 1
return 0 if ans == len(nums) + 1 else ans
```
JavaScript Code：
```js
/*
* @lc app=leetcode id=209 lang=javascript
*
* [209] Minimum Size Subarray Sum
*
*/
/**
* @param {number} s
* @param {number[]} nums
* @return {number}
*/
var minSubArrayLen = function (s, nums) {
if (nums.length === 0) return 0;
const slideWindow = [];
let acc = 0;
let min = null;
for (let i = 0; i < nums.length + 1; i++) {
const num = nums[i];
while (acc >= s) {
if (min === null || slideWindow.length < min) {
min = slideWindow.length;
}
acc = acc - slideWindow.shift();
}
slideWindow.push(num);
acc = slideWindow.reduce((a, b) => a + b, 0);
}
return min || 0;
};
```
C++ Code：
```
class Solution {
public:
int minSubArrayLen(int s, vector& nums) {
int num_len= nums.size();
int left=0, right=0, total=0, min_len= num_len+1;
while (right < num_len) {
do {
total += nums[right++];
} while (right < num_len && total < s);
while (left < right && total - nums[left] >= s) total -= nums[left++];
if (total >=s && min_len > right - left)
min_len = right- left;
}
return min_len <= num_len ? min_len: 0;
}
};
```
**复杂度分析**
* 时间复杂度：$O(N)$，其中 N 为数组大小。
* 空间复杂度：$O(1)$
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### 扩展
如果题目要求是 sum = s, 而不是 sum >= s 呢？
eg:
```js
var minSubArrayLen = function (s, nums) {
if (nums.length === 0) return 0;
const slideWindow = [];
let acc = 0;
let min = null;
for (let i = 0; i < nums.length + 1; i++) {
const num = nums[i];
while (acc > s) {
acc = acc - slideWindow.shift();
}
if (acc === s) {
if (min === null || slideWindow.length < min) {
min = slideWindow.length;
}
slideWindow.shift();
}
slideWindow.push(num);
acc = slideWindow.reduce((a, b) => a + b, 0);
}
return min || 0;
};
```
大家对此有何看法，欢迎给我留言，我有时间都会一一查看回答。更多算法套路可以访问我的 LeetCode 题解仓库： 。 目前已经 37K star 啦。 大家也可以关注我的公众号《力扣加加》带你啃下算法这块硬骨头。 
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