**Leetcode 209. Minimum Size Subarray Sum | 区间和符合条件 = k

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https://leetcode.com/problems/minimum-size-subarray-sum/description/

O(n)可解

class Solution {
public:int minSubArrayLen(int s, vector<int>& nums) {if (nums.size()==0 || s == 0) return 0;int cur = nums[0], left = 0, right = 0;int ret = nums.size() + 1;while (right < nums.size()){if (cur >= s) {ret = min(ret, right - left + 1);cur -= nums[left];left++;} else {right ++;cur += nums[right]; }}        return ret == nums.size() + 1 ? 0 : ret;}
};

另外拿这个题当二分联系了，二分长度

class Solution {
public:int minSubArrayLen(int s, vector<int>& nums) {int cur = 0, left = 0,   l = 0, r = nums.size(), get = 0;for (int i = 0; i < nums.size(); i++) cur += nums[i];if (cur<s) return 0;while (l < r) {int mid ;mid =l + ((r - l )>>1);bool check = false;for (left=0; left + mid -1 < nums.size(); left++) {if (left == 0) {cur = 0;for (int i = left; i < left + mid; i++) {cur += nums[i];}} else {cur -= nums[left-1];cur += nums[left + mid -1];}if (cur >= s) {check = true;get = 1;break;} }if (check) {r = mid;} else {l = mid + 1;}}if (get) return r; else return nums.size();}
};

这个可以作为模板，注意点就是 r=mid这种写法，当长度为1的时候，必须是做端点，对应上面mid的写法。

为了防止死循环，每次代入l=2 r=3检查一次

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