Given an array ofnpositive integers and a positive integers, 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]ands = 7,the subarray[4,3]has the minimal length under the PRoblem constraint.

這道題主要還是先要把題目的要求弄清楚。首先題目中所說的if there is not one, return 0 instead。怎樣才算是沒有呢？

這里意思是如果我們算出來的min length和array本身的長度一樣，那么我們就應該return 0。

這里比較容易想到的是用two pointer來計算。一個pointer用來計算sum，和另一個pointer緊隨其后，用來判斷長度。

代碼如下。~

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;    }}