Minimum Size Subarray Sum
问题:
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.
思路:
双指针
我的代码:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
public class Solution { public int minSubArrayLen(int s, int[] nums) { if(nums==null || nums.length==0) return 0; int minLen = Integer.MAX_VALUE; int len = nums.length; int tmpSum = 0; for(int i=0,j=0; j<len; j++) { tmpSum += nums[j]; if(tmpSum >= s) { while(i <= j) { tmpSum -= nums[i]; if(tmpSum < s) { tmpSum += nums[i]; break; } else i++; } minLen = Math.min(minLen, j-i+1); } } return minLen==Integer.MAX_VALUE ? 0 : minLen; } }
学习之处:
- 双指针的用处在于,后面的指针用于确定下界,前面的指针用于确定上界,然后前面的指针一直再向后指针靠近,尽量压缩之间的距离,也就是要求得的距离
- 明明规划,i 变成start j变成end 更好一点
- 改掉自己的坏习惯,稳住心态,不焦虑