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Commit 7aa9d88

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update: 010
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‎note/010/README.md

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@@ -28,7 +28,7 @@ isMatch("aab", "c*a*b") → true
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## 思路 0
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题意是让让你从判断 `s` 字符串是否正则匹配于 `p`,这道题和[Wildcard Matching][044]很是相似,区别在于 `*`,通配符的 `*` 是可以随意出现的,跟前面字符没有任何关系,其作用是可以表示任意字符串;而正则匹配的 `*` 不能单独存在,前面必须具有一个字符,其意义是表明前面的这个字符个数可以是任意个数,包括 0 个。首先我们用递归的方式来实现,其思路如下:
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题意是让让你从判断 `s` 字符串是否正则匹配于 `p`,这道题和 [Wildcard Matching][044] 很是相似,区别在于 `*`,通配符的 `*` 是可以随意出现的,跟前面字符没有任何关系,其作用是可以表示任意字符串;而正则匹配的 `*` 不能单独存在,前面必须具有一个字符,其意义是表明前面的这个字符个数可以是任意个数,包括 0 个。首先我们用递归的方式来实现,其思路如下:
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* 如果 `s``p` 都为空,那么返回 `true`
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