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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。
s
p
'.'
'*'
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
a-z
.
*
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 }
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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:
示例 2:
示例 3:
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*
。*
时,前面都匹配到有效的字符Solution
Language: JavaScript
The text was updated successfully, but these errors were encountered: