# 1480. Running Sum of 1d Array

#### Easy

***

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`

```python
class Solution:
    def runningSum(self, nums: List[int]) -> List[int]:
        for index in range(len(nums)):
            if index > 0:
                nums[index] = nums[index] + nums[index-1]
        return nums
```