leetcode--Minimum Size Subarray Sum

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.

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More practice:

If you have figured out the O(n) solution, try coding another solution of which the time complexity isO(n log n).

题意：给定一个整数数组，和一个整数s，求某个子数组之和大于这个整数s，且数组长度最短，返回最短长度

如果不存在这样的子数组，返回0

分类：数组，双指针

解法1：双指针思路，一个指针end向后找，直到和大于s,这时计算子数组长度，

然后第二个指针start开始找，和减去第二个指针指向的值，如果这个值还大于s,更新长度，start继续向后

直到和小于s,end才继续找

重复上述过程

[java]  view plain  copy

1. public class Solution {  
2.     public int minSubArrayLen(int s, int[] nums) {  
3.         int start = 0;  
4.         int end = 0;  
5.         int len = nums.length;  
6.         if(len==0) return 0;  
7.         int sum = 0;  
8.         int count = len+1;  
9.         while(start<=end){  
10.             if(sum<s){            <span style="white-space:pre">   </span>  
11.             <span style="white-space:pre">  </span>if(end==len) break;  
12.                 sum += nums[end++];  
13.             }else{  
14.                 count = Math.min(count,end-start);      
15.                 sum -= nums[start++];  
16.             }  
17.         }  
18.         if(count==len+1) return 0;  
19.         return count;  
20.     }  
21. }  

原文链接http://blog.youkuaiyun.com/crazy__chen/article/details/46575663