1480. Running Sum of 1d Array.cpp

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) {
        vector <int> rSum(nums.size(), 0);
        rSum[0] = nums[0];
        for (int i = 1; i < nums.size(); i++) {
            rSum[i] = rSum[i - 1] + nums[i];
        }
        return rSum;
    }
};