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leetcode 10. Regular Expression Matching

2024-10-22 11:54:02   210人阅读

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).

The function prototype should be:
bool isMatch(const char *s, const char *p)

Some examples:
isMatch("aa","a") ? false
isMatch("aa","aa") ? true
isMatch("aaa","aa") ? false
isMatch("aa", "a*") ? true
isMatch("aa", ".*") ? true
isMatch("ab", ".*") ? true
isMatch("aab", "c*a*b") ? true

1.p为空，那s必然也为空

2.s为空，看p的第二字符是否为*，为*的话 则为isMatch(s,p.substring(2));

3.都不为空，p的第二字符为*，一种情况是直接跳过*部分isMatch(s,p.substring(2))，一种情况是首字符相等isMatch(s.substring(1),p)，一种情况是p首字符是.

4.都不为空，p的第二字符不为*，看首字符是否相等

 

 1 public class Solution {
 2     public boolean isMatch(String s, String p) {
 3          if ("".equals(p))
 4             return "".equals(s);
 5         else if ("".equals(s)){
 6             if (p.length()>1&&p.charAt(1)==‘*‘){
 7                 return isMatch(s,p.substring(2));
 8             }else
 9                 return false;
10         }
11         else if (p.length()>1&&p.charAt(1)==‘*‘){
12             if (isMatch(s,p.substring(2)))//skip .*
13                 return true;
14             else if(s.charAt(0)==p.charAt(0)||p.charAt(0)==‘.‘){
15                 return isMatch(s.substring(1),p);
16             }else {
17                 return  false;
18             }
19         }else {
20             if (p.charAt(0)==s.charAt(0)||p.charAt(0)==‘.‘)
21                 return isMatch(s.substring(1),p.substring(1));
22             else
23                 return false;
24         }
25     }
26 }

 

leetcode 10. Regular Expression Matching


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