209. 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). 

 

 

class Solution {
public:
    int minSubArrayLen(int s, vector<int>& nums) {
        int sum=0,i=0,j=0,res=nums.size()+1;
        while(j<nums.size())
        {
            sum+=nums[j++];
            while(sum>=s)
            {
                sum-=nums[i++];
                res=min(res,j-i+1);
            }
        }
        return res>nums.size()?0:res;
    }
};

经典的双指针问题.开始用的是单层for循环, 调试了很久条件总是写的不对,实在不行看了答案, 发现用的是双层循环, 确实这个很简洁, 也是非常直接的逻辑