标签:only 表示 span || har over art 字符 lan
Given an input string (s
) and a pattern (p
), implement regular expression matching with support for ‘.‘
and ‘*‘
.
‘.‘ Matches any single character. ‘*‘ Matches zero or more of the preceding element.
The matching should cover the entire input string (not partial).
Note:
s
could be empty and contains only lowercase letters a-z
.p
could be empty and contains only lowercase letters a-z
, and characters like .
or *
.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 precedeng 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 (.)".
Example 4:
Input: s = "aab" p = "c*a*b" Output: true Explanation: c can be repeated 0 times, a can be repeated 1 time. Therefore it matches "aab".
Example 5:
Input: s = "mississippi" p = "mis*is*p*." Output: false
public boolean isMatch(String text, String pattern) { if (pattern.isEmpty()) return text.isEmpty(); boolean first_match = (!text.isEmpty() && (pattern.charAt(0) == text.charAt(0) || pattern.charAt(0) == ‘.‘)); //只有长度大于 2 的时候,才考虑 * if (pattern.length() >= 2 && pattern.charAt(1) == ‘*‘){ //两种情况 //pattern 直接跳过两个字符。表示 * 前边的字符出现 0 次 //pattern 不变,例如 text = aa ,pattern = a*,第一个 a 匹配,然后 text 的第二个 a 接着和 pattern 的第一个 a 进行匹配。表示 * 用前一个字符替代。 return (isMatch(text, pattern.substring(2)) || (first_match && isMatch(text.substring(1), pattern))); } else { return first_match && isMatch(text.substring(1), pattern.substring(1)); } }
https://leetcode.windliang.cc/leetCode-10-Regular-Expression-Matching.html
厉害了 我的哥
10. Regular Expression Matching
标签:only 表示 span || har over art 字符 lan
原文地址:https://www.cnblogs.com/wentiliangkaihua/p/10353662.html