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###### tags: `Leetcode`
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# Leetcode 1480. Running Sum of 1d Array

[link](https://leetcode.com/problems/running-sum-of-1d-array/)

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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

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題意: 給定一個陣列，回傳一個「累積和」陣列

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We can iterate the array and the i item is the sum of the i-1 item + the original i item. Notice that the first element is simply the original first item.

#### Solution 1

class Solution:
    def runningSum(self, nums: List[int]) -> List[int]:
        lst = []
        for i in range(len(nums)):
            if i == 0 :
                lst.append(nums[i])
            else:
                lst.append(nums[i] + lst[i - 1])

        return lst
        

O(T): O(n)
O(S): O(n)

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Directly update the original array.

#### Solution 2

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

        return nums

O(T): O(n)
O(S): O(1)