209. 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.
思路:设置一个i和j，初始值都为0，然后j一直往后走，知道sum[i...j]>=s了，然后将i向前走，直到第一个sum[i..j]<s

下面举一个例子

2 3 1 2 4 3

i =0 ，j=0； minlen=Integer.Max_Value

i=0，j=3；minlen=4；

i=1 ,j=3;minlen=4;

i=1,j=4;minlen=4;

i=3,j=4;minlen=3;

i=3;j=5;,minlen=3;

所以minlen=3；

代码如下(已通过leetcode)

public class Solution {

public int minSubArrayLen(int s, int[] nums) {

if(nums==null ||nums.length==0) return 0;

int i=0,j=0;

int sum=nums[0];

int minlen=Integer.MAX_VALUE;

while(i<nums.length&&j<nums.length) {

if(sum<s) {

j++;

if(j>=nums.length) break;

else sum=sum+nums[j];

}

else{

minlen=Math.min(minlen, j-i+1);

//System.out.println(minlen);

sum=sum-nums[i];

i++;

}

}

if(minlen==Integer.MAX_VALUE) return 0;

else return minlen;

}

}