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小墨
2022-05-02 23:44:12 +08:00
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leetcode-cn/originData/find-triangular-sum-of-an-array.json
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@@ -2,7 +2,7 @@
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"questionId": "2324",
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"categoryTitle": "Algorithms",
"boundTopicId": 1385552,
"title": "Find Triangular Sum of an Array",
@@ -12,13 +12,38 @@
"translatedContent": "<p>给你一个下标从 <strong>0</strong>&nbsp;开始的整数数组&nbsp;<code>nums</code>&nbsp;，其中&nbsp;<code>nums[i]</code>&nbsp;是 <code>0</code>&nbsp;到 <code>9</code>&nbsp;之间（两者都包含）的一个数字。</p>\n\n<p><code>nums</code>&nbsp;的 <strong>三角和</strong>&nbsp;是执行以下操作以后最后剩下元素的值：</p>\n\n<ol>\n\t<li><code>nums</code>&nbsp;初始包含&nbsp;<code>n</code>&nbsp;个元素。如果&nbsp;<code>n == 1</code>&nbsp;，<strong>终止</strong>&nbsp;操作。否则，<strong>创建</strong>&nbsp;一个新的下标从&nbsp;<strong>0</strong>&nbsp;开始的长度为 <code>n - 1</code>&nbsp;的整数数组&nbsp;<code>newNums</code>&nbsp;。</li>\n\t<li>对于满足&nbsp;<code>0 &lt;= i &lt;&nbsp;n - 1</code>&nbsp;的下标&nbsp;<code>i</code>&nbsp;，<code>newNums[i]</code> <strong>赋值</strong>&nbsp;为&nbsp;<code>(nums[i] + nums[i+1]) % 10</code>&nbsp;，<code>%</code>&nbsp;表示取余运算。</li>\n\t<li>将&nbsp;<code>newNums</code>&nbsp;<strong>替换</strong> 数组&nbsp;<code>nums</code>&nbsp;。</li>\n\t<li>从步骤 1 开始&nbsp;<strong>重复</strong>&nbsp;整个过程。</li>\n</ol>\n\n<p>请你返回&nbsp;<code>nums</code>&nbsp;的三角和。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<p><img alt=\"\" src=\"https://assets.leetcode.com/uploads/2022/02/22/ex1drawio.png\" style=\"width: 250px; height: 250px;\" /></p>\n\n<pre>\n<b>输入：</b>nums = [1,2,3,4,5]\n<b>输出：</b>8\n<strong>解释：</strong>\n上图展示了得到数组三角和的过程。</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<b>输入：</b>nums = [5]\n<b>输出：</b>5\n<b>解释：</b>\n由于 nums 中只有一个元素，数组的三角和为这个元素自己。</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= nums.length &lt;= 1000</code></li>\n\t<li><code>0 &lt;= nums[i] &lt;= 9</code></li>\n</ul>\n",
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"difficulty": "Medium",
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"translatedName": "数组",
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@@ -130,7 +155,7 @@
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"hints": [
"Try simulating the entire process.",
"To reduce space, use a temporary array to update nums in every step instead of creating a new array at each step."