No1480.running-sum-of-1d-array.cs

/*
 * Difficulty:
 * Easy
 *
 * Desc:
 * 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
*/

public class Solution {
    public int[] RunningSum(int[] nums) {
        return nums.Aggregate(new List<int>(new int[]{ 0 }), (list, num) => {
            list.Add(list[^1] + num);
            return list;
        }).ToArray()[1..];
    }
}