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.
这道题主要还是先要把题目的要求弄清楚。首先题目中所说的if there is not one, return 0 instead。怎样才算是没有呢？
这里意思是如果我们算出来的min length和array本身的长度一样，那么我们就应该return 0。
这里比较容易想到的是用two pointer来计算。一个pointer用来计算sum，和另一个pointer紧随其后，用来判断长度。
代码如下。~
public class Solution {
public int minSubArrayLen(int s, int[] nums) {
//two pointer
//one from start;one follows
int start = 0;
int end = 0;
int sum = 0;
int min = nums.length;
while(start<nums.length && end<nums.length) {
while(sum<s && end<nums.length) {
sum=sum+nums[end];
end++;
}
while(sum>=s && start<=end) {
min = Math.min(min, end-start);
sum =sum-nums[start];
start++;
}
}
if(min==nums.length){
return 0;
}
return min;
}
}