LeetCode 筆記｜1480. Running Sum of 1d Array

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計算數列逐項的累積和 (runningSum)

問題描述

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.

樣例

Input: nums = [1,2,3,4]
Output: [1,3,6,10]

限制

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

知識點
prefix sum & array

思路與題解

此題應該算是 for loop 練習，我覺得要留意 (1) indices out of range 以及 (2) len(nums) == 1 的狀況。

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class Solution:
    def runningSum(self, nums: List[int]) -> List[int]:
        for n in range(1, len(nums)):
            nums[n] = nums[n - 1] + nums[n]
        return(nums)