题目顺序：代码随想录算法公开课，b站上有相应视频讲解

一、题目

209. Minimum Size Subarray Sum

Given an array of positive integers nums and a positive integer target, return the minimal length of a
subarray
whose sum is greater than or equal to target. If there is no such subarray, return 0 instead.

Example 1:

Input: target = 7, nums = [2,3,1,2,4,3]
Output: 2
Explanation: The subarray [4,3] has the minimal length under the problem constraint.
Example 2:

Input: target = 4, nums = [1,4,4]
Output: 1
Example 3:

Input: target = 11, nums = [1,1,1,1,1,1,1,1]
Output: 0

Constraints:

1 <= target <= 109
1 <= nums.length <= 105
1 <= nums[i] <= 104

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)).

二、题解

双指针滑动窗口的思路

j代表滑动窗口末尾的索引，i代表初始索引

class Solution {
public:
    int minSubArrayLen(int target, vector<int>& nums) {
        int n = nums.size();
        int i = 0,sum = 0,len = n + 1;
        for(int j = 0;j < n;j++){
            sum += nums[j];
            while(sum >= target){
                len = min(len,j - i + 1);
                sum -= nums[i];
                i++;
            }
        }
        return len == n + 1 ? 0 : len;
    }
};