# 【LeetCode】 1480. Running Sum of 1d Array

## Description
> 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`.

> 給予一個陣列 `nums`。 我們定義一個總和陣列為 `runningSum[i] = sum(nums[0]…nums[i])`。
> 回傳 `nums` 的總和陣列。

## Example:

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]

## Solution

* 可以再創一個陣列，然後真的去記錄總和並放到陣列之中
* 更快的方法是直接把陣列中的每一個元素加上前一個元素即可

### Code

class Solution {
public:
    vector<int> runningSum(vector<int>& nums) {
        for(int i = 1; i < nums.size(); i++)
            nums[i] += nums[i - 1];
        return nums;
    }
};

###### tags: `LeetCode` `C++`