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.
Example:
Input: s = 7, nums = [2,3,1,2,4,3] Output: 2 Explanation: the subarray [4,3] has the minimal length under the problem constraint.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).
题目大意：
题目中给出一个目标，和一个数组，求出长度最少的区间满足和>=s。
解题思路：
从第一个位置开始遍历，当找到满足要求时将左标签l向右移动。保存最小距离，再将右标签r向右移动。
class Solution { public: int minSubArrayLen(int s, vector<int>& nums) { // O(n) l and r int n = nums.size(); int ans = INT_MAX; int l = 0, r = 0; int sum = 0; while(r<n){ sum += nums[r]; while(sum>=s){ ans = min(ans, r+1-l); sum -= nums[l++]; } r++; } return (ans==INT_MAX)?0:ans; } };
