1480. Running Sum of 1d Array

Difficulty:
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    Problem

    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:

    Solution (Java)

    class Solution {
        public int[] runningSum(int[] nums) {
            int sum = 0;
            int[] result = new int[nums.length];
            for (int i = 0; i < nums.length; i++) {
                sum += nums[i];
                result[i] = sum;
            }
            return result;
        }
    }
    

    Explain:

    nope.

    Complexity: