Problem:

Given an array of n positive integers and a positive integer s, find the minimal length of a subarray of which the sum ≥ s. If there isn't one, return 0 instead.

For example, given the array [2,3,1,2,4,3] and s = 7,
the subarray [4,3] has the minimal length under the problem constraint.

Summary:

找到数组中和大于目标值s的最短子序列。

Solution:

用两个指针分别代表当前子序列的开始和结尾,若计算出的sum小于s,则end++,否则start--,每计算出一个更短的子序列,更新res值。

 class Solution {
public:
int minSubArrayLen(int s, vector<int>& nums) {
int len = nums.size();
int res = len + , start = , end = , sum = ;
while (start < len && end < len) {
while (sum < s && end < len) {
sum += nums[end++];
} while (sum >= s && start <= end) {
res = min(res, end - start);
sum -= nums[start++];
}
} return res == len + ? : res;
}
};

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