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.
Credits:
Special thanks to @Freezen for adding this problem and creating all test cases.
双指针,用cmp保存数列的值,每次判断(前一保存数组与当前值的和) 与 当index前值的大小,如果小于当前index的值,说明前段数组小于零舍弃,从当前index计数。如果cmp大于s,则从left指针开始舍弃,直到cmp小于s,保存数组的长度,具体的代码如下
public class Solution { public int minSubArrayLen(int s, int[] nums) { if(nums.length<=0||s<=0) return 0; int cmp = nums[0], res = 1,minRes = Integer.MAX_VALUE; int left = 0; if(cmp>=s) return res; for(int i=1;i<nums.length;i++){ if(nums[i]>=cmp +nums[i]){ left = i; cmp = nums[i]; res = 1; }else{ cmp = cmp+nums[i]; res++; } if(cmp>=s){ while(cmp>=s){ cmp -= nums[left]; left++; res--; } minRes = Math.min(res+1, minRes); } } return minRes==Integer.MAX_VALUE?0:minRes; } }
原文地址:http://blog.csdn.net/guorudi/article/details/45750427