---
title: Leetcode 1480. Running Sum of 1d Array
tags: leetcode
---

# 1480. Running Sum of 1d Array

## 題目連結

https://leetcode.com/problems/running-sum-of-1d-array/description/

## 題目敘述

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.

> return的array[i]是原本array[0]+...+原本array[i]

**Example 1**

:::success
Input: nums = [1,2,3,4]
Output: [1,3,6,10]
Explanation: [1, 1+2, 1+2+3, 1+2+3+4].
:::

**Example 2**

:::success
Input: nums = [1,1,1,1,1]
Output: [1,2,3,4,5]
:::

**Example 3**

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

**Constraints**

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

## 解題思路

原本想說透過兩層迴圈去，最外層跑遍nums[0]到nums[nums.length-1]，內層從nums[0]到nums[i]，i為目前跑到的index，然後存入新的陣列然後回傳。
後來透過 [Solutions](https://leetcode.com/problems/running-sum-of-1d-array/solutions/3167663/simple-solution-java-cpp-python-c-language-runtime-1-ms-beats-100/) 發現用一層就可以了ㄟ，nums中每個元素的值就等於目前的元素加上前一項：`nums[i] += nums[i-1]`，由於前一項會是前面元素加起來的合，因此透過一個迴圈跑即可。

**java程式碼**

class Solution {
    public int[] runningSum(int[] nums) {
        for(int i = 1 ; i < nums.length ; i++) {
            nums[i] += nums[i-1];
        }
        return nums;
    }
}

**執行結果**