每天一算：Minimum Size Subarray Sum

 1// 209. Minimum Size Subarray Sum
 2// https://leetcode.com/problems/minimum-size-subarray-sum/description/
 3//
 4// 滑动窗口的思路
 5// 时间复杂度: O(n)
 6// 空间复杂度: O(1)
 7class Solution {
 8public:
 9    int minSubArrayLen(int s, vector<int>& nums) {
10
11        int l = 0 , r = -1; // nums[l...r]为我们的滑动窗口
12        int sum = 0;
13        int res = nums.size() + 1;
14
15        while(l < nums.size()){   // 窗口的左边界在数组范围内,则循环继续
16
17            if(r + 1 < nums.size() && sum < s)
18                sum += nums[++r];
19            else // r已经到头 或者 sum >= s
20                sum -= nums[l++];
21
22            if(sum >= s)
23                res = min(res, r - l + 1);
24        }
25
26        if(res == nums.size() + 1)
27            return 0;
28        return res;
29    }
30};