Running-Sum-of-1d-Array

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Algorithms

Given an array <code>nums</code>. We define a running sum of an array as&nbsp;<code>runningSum[i] = sum(nums[0]&hellip;nums[i])</code>.

Return the running sum of <code>nums</code>.

&nbsp;
Example 1:

<pre>
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].</pre>

Example 2:

<pre>
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].</pre>

Example 3:

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

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= nums.length &lt;= 1000</code></li>
	<li><code>-10^6&nbsp;&lt;= nums[i] &lt;=&nbsp;10^6</code></li>
</ul>

runningSum

Running Sum of 1d Array

running-sum-of-1d-array

Array

Prefix Sum

给你一个数组 <code>nums</code> 。数组「动态和」的计算公式为：<code>runningSum[i] = sum(nums[0]&hellip;nums[i])</code> 。

请返回 <code>nums</code> 的动态和。

&nbsp;

示例 1：

<pre>输入：nums = [1,2,3,4]
输出：[1,3,6,10]
解释：动态和计算过程为 [1, 1+2, 1+2+3, 1+2+3+4] 。</pre>

示例 2：

<pre>输入：nums = [1,1,1,1,1]
输出：[1,2,3,4,5]
解释：动态和计算过程为 [1, 1+1, 1+1+1, 1+1+1+1, 1+1+1+1+1] 。</pre>

示例 3：

<pre>输入：nums = [3,1,2,10,1]
输出：[3,4,6,16,17]
</pre>

&nbsp;

提示：

<ul>
	<li><code>1 &lt;= nums.length &lt;= 1000</code></li>
	<li><code>-10^6&nbsp;&lt;= nums[i] &lt;=&nbsp;10^6</code></li>
</ul>