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Mar 23, 2022
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docs: update a description of the solution to lc problem: No.0049 #768

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8 changes: 6 additions & 2 deletions solution/0000-0099/0049.Group Anagrams/README.md
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Expand Up @@ -45,15 +45,19 @@

<!-- 这里可写通用的实现逻辑 -->

遍历字符串，将每个字符串按照字符字典序排序后得到一个新的字符串，将相同的新字符串放在哈希表的同一个 key 对应 value 列表中。
1. 遍历字符串，对每个字符串按照**字符字典序**排序，得到一个新的字符串。
2. 以新字符串为 `key`，`[str]` 为 `value`，存入哈希表当中（`HashMap<String, List<String>>`）。
3. 后续遍历得到相同 `key` 时，将其加入到对应的 `value` 当中即可。

以 `strs = ["eat", "tea", "tan", "ate", "nat", "bat"]` 为例，遍历结束时，哈希表的状况：

| key | value |
| ------- | ----------------------- |
| `"aet"` | `["eat", "tea", "ate"]` |
| `"ant"` | `["tan", "nat"] ` |
| `"abt"` | `["bat"] ` |

最后返回哈希表的 value 列表即可。
最后返回哈希表的 `value` 列表即可。

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