1480. Running Sum of 1d Array

array

https://leetcode.com/problems/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
  • code
class Solution:
    def runningSum(self, nums: List[int]) -> List[int]:
        return list(accumulate(nums))