1480. Running Sum of 1d Array

题目

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

func runningSum(nums []int) []int {
	result := []int{}
	counter := 0

	for x := 0; x < len(nums); x++ {
		for y := 0; y < x; y++ {
			counter += nums[y]
		}

		val := counter + nums[x]
		result = append(result, val)

		counter = 0
	}

	return result
}