maximum average subarray i

Input: [1,12,-5,-6,50,3], k = 4
Output: 12.75
Explanation: Maximum average is (12-5-6+50)/4 = 51/4 = 12.75

1. 1 <= k <= n <= 30,000.
2. Elements of the given array will be in the range [-10,000, 10,000].

class Solution {
public:
    double findMaxAverage(vector<int>& nums, int k) {
        double m = numeric_limits<double>::lowest();
        for (int i = 0; i < nums.size() - k + 1; i++)
        {
            double t = nums[i];
            for (int j = 1; j < k; j++)
            {
                t += nums[i + j];
            }
            m = max(m, t / k);
        }
        return m;
    }
};

class Solution {
public:
    double findMaxAverage(vector<int>& nums, int k) {
        vector<int> sum;
        sum.push_back(nums[0]);
        for (int i = 1; i < nums.size(); i++) {
            sum.push_back(nums[i] + sum[i - 1]);
        }
        double m = sum[k - 1] * 1.0 / k;
        for (int i = k; i < nums.size(); i++) {
            m = max(m, (sum[i] - sum[i - k]) * 1.0 / k);
        }
        return m;
    }
};