LeetCode Q209 Minimum Size Subarray Sum

Question:
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.
Solution:
Solution 1: brutal force + pruning — 239ms

class Solution {
public:
//solution 1 brutal force
    int minSubArrayLen( int s, vector<int>& nums ) {
        int n = nums.size(), len = n + 1;
        for( int i = 0; i < n; ++i ){
            int sum = 0, localLen = 0, j;
            for(j = i; j < nums.size(); ++j ){
                sum += nums[j];
                localLen++;
                if ( sum >= s || localLen == len ){
                    break;
                }
            }
            if (j < nums.size())
                len = min(localLen, len);
            else if ( sum < s )
                break;
        }
        return ( len > n ) ? 0 : len;
    }
};

Solution 2: O(n) –8ms

class Solution {
public:
//Solution 2: using sliding window O(n) -- 8ms
    int minSubArrayLen(int s, vector<int>& nums) {
        int n = nums.size();
        int end = 0, start = 0, sum = 0, len = INT_MAX;
        while( end < n ){
            //find the sliding window
            while( end < n && sum < s )
                sum += nums[end++];
            //find it change the the sliding window's start point
            while( sum >= s && start < n){
                len = min( len, end - start );
                sum -= nums[start++];
            }
        }
        return (len > n ) ? 0 : len;
    }
};