leetcode209. Minimum Size Subarray Sum

Given an array of n positive integers and a positive integer s, find the minimal length of a contiguous subarray of which the sum ≥ s. If there isn't one, return 0 instead.

Example: 

Input: s = 7, nums = [2,3,1,2,4,3]
Output: 2
Explanation: the subarray [4,3] has the minimal length under the problem constraint.

Follow up:

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

思路： 双索引技术；维护一个滑动窗口  slip window [left，right]；表示的是从left到right的数组；能够满足条件；

CODE:

class Solution {
public:
    int minSubArrayLen(int s, vector<int>& nums) 
    {
        int n=nums.size();
        if(n==0)
            return 0;
        int left=0;int right=-1;//[left,right]表示当前的滑动窗口的大小 也是我们需要维护的量这个刚开始的时候是没有值的 所以 right=-1；
        int sum=0;
        int res=n+1;//初始化为不可能取到的值；
        while(left<n)
        {
            if(right+1<n&&sum<s)//注意这里有right的取值 一定不能够忘记  否则会数组越界；
                sum+=nums[++right];
            else 
                sum-=nums[left++];
            if(sum>=s)
                res=min(res,right+1-left);
        }
        return res==n+1?0:res;
    }
};