题目：
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.
If you have figured out the O(n) solution, try coding another solution of which the time complexity is O(n log n).
思路：
两个指针, start end, end向后走,直到 sum 大于 s. 然后start向后, 直到sum 小于s. 同时更新 min值。
/** * @param {number} s * @param {number[]} nums * @return {number} */ var minSubArrayLen = function(s, nums) { var l=0,r=0,min=2147483647,sum=0; while(r<nums.length&&l<=r){ while(r<nums.length&&sum<s){ sum+=nums[r++]; } while(l<=r&&sum>=s){ min=Math.min(min,r-l); sum-=nums[l++]; } } return min==2147483647?0:min; };
原文：http://www.cnblogs.com/shytong/p/5115059.html