10.regular-expression-matching
Statement
Metadata
- Link: 正则表达式匹配
- Difficulty: Hard
- Tag:
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 <= 201 <= p.length <= 30s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
Metadata
- Link: Regular Expression Matching
- Difficulty: Hard
- Tag:
RecursionStringDynamic Programming
Given an input string s and a pattern p, implement regular expression matching with support for '.' and '*' where:
'.'Matches any single character.'*'Matches zero or more of the preceding element.
The matching should cover the entire input string (not partial).
Example 1:
Input: s = "aa", p = "a"
Output: false
Explanation: "a" does not match the entire string "aa".
Example 2:
Input: s = "aa", p = "a*"
Output: true
Explanation: '*' means zero or more of the preceding element, 'a'. Therefore, by repeating 'a' once, it becomes "aa".
Example 3:
Input: s = "ab", p = ".*"
Output: true
Explanation: ".*" means "zero or more (*) of any character (.)".
Constraints:
1 <= s.length <= 201 <= p.length <= 30scontains only lowercase English letters.pcontains only lowercase English letters,'.', and'*'.- It is guaranteed for each appearance of the character
'*', there will be a previous valid character to match.
最后更新: October 11, 2023