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.
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).
好题目，想法有了，就是实现的时候出现了偏差，导致很长时间没有ac，后来还是看的别人的代码
class Solution { public: int minSubArrayLen(int s, vector<int>& nums) { int n = nums.size(); if (n == 0) return 0; int l = 0, r = 0; int sum = 0; int Min = n + 1; while (l < n && r < n) { while (sum < s && r < n) sum += nums[r++]; while (sum >= s && l < n) { Min = min(Min, r - l); sum -= nums[l++]; } } if (Min == n + 1) return 0; return Min; } };
leetcode 209. Minimum Size Subarray Sum
原文：http://www.cnblogs.com/pk28/p/7281295.html