<p>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>.</p>

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

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,3,4]
<strong>Output:</strong> [1,3,6,10]
<strong>Explanation:</strong> Running sum is obtained as follows: [1, 1+2, 1+2+3, 1+2+3+4].</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,1,1,1,1]
<strong>Output:</strong> [1,2,3,4,5]
<strong>Explanation:</strong> Running sum is obtained as follows: [1, 1+1, 1+1+1, 1+1+1+1, 1+1+1+1+1].</pre>

<p><strong class="example">Example 3:</strong></p>

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

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<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>

class Solution {
public:
    vector<int> runningSum(vector<int>& nums) {
        vector<int> result;
        result.reserve(nums.size());

        result.push_back(nums[0]);
        for (int i = 1; i < nums.size(); ++i) {
            result.push_back(result[i - 1] + nums[i]);
        }

        return result;
    }
};
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