Running Sum of 1d Array

Given an array nums. We define a running sum of an array as runningSum[i] = sum(nums[0]…nums[i]).

Return the running sum of nums.

Example 1:
Input: nums = [1,2,3,4]
Output: [1,3,6,10]
Explanation: Running sum is obtained as follows: [1, 1+2, 1+2+3, 1+2+3+4].

Example 2:
Input: nums = [1,1,1,1,1]
Output: [1,2,3,4,5]
Explanation: Running sum is obtained as follows: [1, 1+1, 1+1+1, 1+1+1+1, 1+1+1+1+1].

Example 3:
Input: nums = [3,1,2,10,1]
Output: [3,4,6,16,17]

Constraints:
1 <= nums.length <= 1000
-10^6 <= nums[i] <= 10^6

class Solution 
{
public:
    vector<int> runningSum(vector<int>& nums) 
    {
        
        int count = nums.size();
        vector<int> result_sum(count);
        int before_sum = 0;
        
        for (int i = 0; i < count; ++i)
        {
            result_sum[i] = before_sum + nums[i];
            before_sum = result_sum[i];
        }
        
        
        return result_sum;
    }
};