feat(leecode): 49.字母异位词分组

docs/10.LeeCode/48.旋转图像.md

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 <p>&nbsp;</p>
 
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-<<< @/src/LeeCode/48.旋转图像 1.js
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docs/10.LeeCode/49.字母异位词分组.md

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+---
+title: 49.字母异位词分组
+date: 2024-03-23
+lang: 'zh-CN'
+sidebar: 'auto'
+categories:
+  - LeeCode
+tags:
+location: HangZhou
+---
+
+# Heading
+
+[[toc]]
+
+[49.字母异位词分组](https://leetcode.cn/problems/group-anagrams/description/)
+
+Tags: algorithms amazon bloomberg facebook uber yelp hash-table string
+
+Langs: c cpp csharp dart elixir erlang golang java javascript kotlin php python python3 racket ruby rust scala swift typescript
+
+- algorithms
+- Medium (67.90%)
+- Likes: 1857
+- Dislikes: -
+- Total Accepted: 657.2K
+- Total Submissions: 967.4K
+- Testcase Example: '["eat","tea","tan","ate","nat","bat"]'
+
+<p>给你一个字符串数组，请你将 <strong>字母异位词</strong> 组合在一起。可以按任意顺序返回结果列表。</p>
+
+<p><strong>字母异位词</strong> 是由重新排列源单词的所有字母得到的一个新单词。</p>
+
+<p>&nbsp;</p>
+
+<p><strong>示例 1:</strong></p>
+
+<pre>
+<strong>输入:</strong> strs = <code>["eat", "tea", "tan", "ate", "nat", "bat"]</code>
+<strong>输出: </strong>[["bat"],["nat","tan"],["ate","eat","tea"]]</pre>
+
+<p><strong>示例 2:</strong></p>
+
+<pre>
+<strong>输入:</strong> strs = <code>[""]</code>
+<strong>输出: </strong>[[""]]
+</pre>
+
+<p><strong>示例 3:</strong></p>
+
+<pre>
+<strong>输入:</strong> strs = <code>["a"]</code>
+<strong>输出: </strong>[["a"]]</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= strs.length &lt;= 10<sup>4</sup></code></li>
+	<li><code>0 &lt;= strs[i].length &lt;= 100</code></li>
+	<li><code>strs[i]</code>&nbsp;仅包含小写字母</li>
+</ul>
+
+<<< @/src/LeeCode/49.字母异位词分组.js

src/LeeCode/49.字母异位词分组.js

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+/*
+ * @lc app=leetcode.cn id=49 lang=javascript
+ *
+ * [49] 字母异位词分组
+ */
+
+// @lc code=start
+/**
+ * @param {string[]} strs
+ * @return {string[][]}
+ */
+var groupAnagrams = function (strs) {
+  const map = new Map()
+
+  for (let str of strs) {
+    const key = str.split('').sort().join()
+    map.set(key, map.has(key) ? [...map.get(key), str] : [str])
+  }
+
+  return [...map.values()]
+}
+// @lc code=end