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10. 正则表达式匹配 #6

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webVueBlog opened this issue Sep 2, 2022 · 1 comment
Open

10. 正则表达式匹配 #6

webVueBlog opened this issue Sep 2, 2022 · 1 comment

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@webVueBlog
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10. 正则表达式匹配

Description

Difficulty: 困难

Related Topics: 递归, 字符串, 动态规划

给你一个字符串 s 和一个字符规律 p,请你来实现一个支持 '.' 和 '*' 的正则表达式匹配。

  • '.' 匹配任意单个字符
  • '*' 匹配零个或多个前面的那一个元素

所谓匹配,是要涵盖 **整个 **字符串 s的,而不是部分字符串。

示例 1:

输入:s = "aa", p = "a"
输出:false
解释:"a" 无法匹配 "aa" 整个字符串。

示例 2:

输入:s = "aa", p = "a*"
输出:true
解释:因为 '*' 代表可以匹配零个或多个前面的那一个元素, 在这里前面的元素就是 'a'。因此,字符串 "aa" 可被视为 'a' 重复了一次。

示例 3:

输入:s = "ab", p = ".*"
输出:true
解释:".*" 表示可匹配零个或多个('*')任意字符('.')。

提示:

  • 1 <= s.length <= 20
  • 1 <= p.length <= 30
  • s 只包含从 a-z 的小写字母。
  • p 只包含从 a-z 的小写字母,以及字符 . 和 *
  • 保证每次出现字符 * 时,前面都匹配到有效的字符

Solution

Language: JavaScript

/**
 * @param {string} s
 * @param {string} p
 * @return {boolean}
 */
var isMatch = function(s, p) {
    p = new RegExp(p)
    s = s.replace(p, '')
    return s === '' ? true : false
}
@suedar
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suedar commented Feb 10, 2023

这难度困难,你在逗我

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