【滑动窗口】最小长度连续子序列和的大小判定

/LeetCode :209Description:Given an array of n positive integers and a positive integer s, find the minimal length of a contiguous subarray of which the sum ≥ s. If there isn’t one, return 0 instead.
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). /#includeusing namespace std;/思路：双指针法/滑动窗口法1. 从第0项开始不断累加 sum2. 一旦 sum 超过 s,记录当前子序列长度3. 不断剔除当前子序列的首项，直到它小于s
notes:窗口是什么？：满足sum>=s的最小长度连续子数组窗口如何移动：若当前窗口值大于等于s,即需要缩小窗口的结束位置：即遍历数组的指针，起始位置即数组的起始位置时间复杂度：O(n) 空间复杂度:O(1)/class Solution {public: int minSubArrayLen(int s, vector& nums) { int result = INT32_MAX; int slowPtr = 0; int sum = 0; int curLength = 0; for (int fastPtr = 0; fastPtr < nums.size(); fastPtr++) { sum += nums[fastPtr]; while (sum>=s) { curLength = fastPtr - slowPtr + 1; result = curLength <= result ? curLength : result; sum -= nums[slowPtr++]; } } return result == INT32_MAX ? 0 : result; }};
int main() { Solution ss; vector nums = {2, 3, 1, 2, 4, 3}; cout << ss.minSubArrayLen(7, nums) << endl; return 0;}