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).
class Solution { public int minSubArrayLen(int s, int[] nums) { if(nums.length == 0 || nums == null) return 0; int n = nums.length; int left = 0, sum = 0, res = Integer.MAX_VALUE; for(int i=0; i<n; i++){ sum += nums[i]; while(sum >= s){ res = Math.min(res, i-left+1); sum -= nums[left++]; } } return res == Integer.MAX_VALUE ? 0 : res; } }转载于:https://www.cnblogs.com/Roni-i/p/11222485.html
原文链接：https://blog.csdn.net/weixin_30342827/article/details/99664771
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