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leetcode-problemset/leetcode-cn/problem (Chinese)/一维数组的动态和 [running-sum-of-1d-array].html

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2022-03-27 20:37:52 +08:00
<p>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></p>
<p>请返回 <code>nums</code> 的动态和。</p>
<p>&nbsp;</p>
<p><strong>示例 1</strong></p>
<pre><strong>输入:</strong>nums = [1,2,3,4]
<strong>输出:</strong>[1,3,6,10]
<strong>解释:</strong>动态和计算过程为 [1, 1+2, 1+2+3, 1+2+3+4] 。</pre>
<p><strong>示例 2</strong></p>
<pre><strong>输入:</strong>nums = [1,1,1,1,1]
<strong>输出:</strong>[1,2,3,4,5]
<strong>解释:</strong>动态和计算过程为 [1, 1+1, 1+1+1, 1+1+1+1, 1+1+1+1+1] 。</pre>
<p><strong>示例 3</strong></p>
<pre><strong>输入:</strong>nums = [3,1,2,10,1]
<strong>输出:</strong>[3,4,6,16,17]
</pre>
<p>&nbsp;</p>
<p><strong>提示:</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>