# LeetCode 1480

###### tags: `python`,`LeetCode`
>這邊使用Python解題
## 題目：
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.

## 範例
### 範例 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].

### 範例 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].

### 範例 3:

Input: nums = [3,1,2,10,1]
Output: [3,4,6,16,17]

## 條件限制
* 1 <= nums.length <= 1000
* -10^6 <= nums[i] <= 10^6

## 我的解題思路：
這題其實是累加，將每個元素與目前狀態做遞加而已

ans[0] = nums[0]
ans[1] = nums[0+1]
ans[2] = nums[0+1+2]

1. need to know the current state/position => index
2. iterate to the current position in the nums array

## 程式碼:

def runningSum(nums):
    for i in range(1, len(nums)):
        nums[i] += nums[i-1]
    return nums

## 測試

numss = [[1,2,3,4], [1,1,1,1,1], [3,1,2,10,1]]
anss = [[1,3,6,10], [1,2,3,4,5], [3,4,6,16,17]]
for idx in range(len(numss)):
    result = None
    try:
        result = runningSum(numss[idx])
    except Exception as ex:
        print(ex)
    if result == anss[idx]:
        print('case 1: pass')
    else:
        print('case 1: failed. your ans is', result, 'expect', anss[idx])