LeetCode—Minimum Size Subarray Sum

题目：

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.

代码1：遍历，复杂度O(n^2)

int min(int x,int y)
{
    return x<y?x:y;
}
int minSubArrayLen(int s, int* nums, int numsSize)
{
    int minlen = 100000;
    int count = 0;
    int i;
    int j;
    for(i = 0;i < numsSize;i ++)
    {
            count+=nums[i];
    }
    if(count<s)
    return 0;
    count = 0;
    for(i = 0;i<numsSize;i++)
    {
            count = 0;
            for(j = i;j>=0;j--)
            {
                count += nums[j];
                if(s <= count)
                {
                   minlen = min(minlen,i-j+1);
                   break;
                }
            }
    }    
    return minlen;
}

C代码2：滑动窗口，复杂度O(n)

int minSubArrayLen(int s, int* nums, int numsSize)
{
    int minlen = 100000;
    int left  = 0;
    int right = 0;
    int sum = 0;
    while (right <= numsSize) 
     {
        if (sum >= s) 
            sum -= nums[left++];
        else 
            sum += nums[right++];
        if (right - left < minlen && sum >= s)
            minlen = right - left;
        if(right == numsSize && sum < s)
            break;
     }
     return minlen == 100000 ? 0 : minlen;
}

复杂度为O(n*logn)的没想出来，thinking。。。