LeetCode Brush Questions 1480. Dynamic Sum of One-Dimensional Array

LeetCode Brush Questions 1480. Dynamic Sum of One-Dimensional Array

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• Subject :
  Here is an array for you nums. Array formula "and dynamic" as follows: runningSum[i] = sum(nums[0]…nums[i]). Please return to numsthe dynamic and.
• Example :

示例 1 :
输入：nums = [1,2,3,4]
输出：[1,3,6,10]
解释：动态和计算过程为 [1, 1+2, 1+2+3, 1+2+3+4] 。

示例 2 :
输入：nums = [1,1,1,1,1]
输出：[1,2,3,4,5]
解释：动态和计算过程为 [1, 1+1, 1+1+1, 1+1+1+1, 1+1+1+1+1] 。

示例 3:
输入：nums = [3,1,2,10,1]
输出：[3,4,6,16,17]

• Tips :
  • 1 <= nums.length <= 1000
  • -10^6 <= nums[i] <= 10^6
• Code:

class Solution:
    def runningSum(self, nums: List[int]) -> List[int]:
        result = [0]
        for i in range(len(nums)):
            result.append(result[-1] + nums[i])
        return result[1:]
# 执行用时：36 ms, 在所有 Python3 提交中击败了91.69%的用户
# 内存消耗：13.9 MB, 在所有 Python3 提交中击败了5.30%的用户

• Algorithm description: It is
  not possible to sum up from the beginning every time, resultadd new elements to the last result used , and add it to resultit.