【leetcode刷题笔记】Regular Expression Matching

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

• s为0的时候，如果p为1，false；否则如果p的第1位上为‘*‘，那么就要考察p后面的元素，于是递归调用 isMatch(s, p.substring(2)) ，这样的例子有s = "", p = "c*c*"；如果p的第一位上不为1，那么s和p肯定不匹配了。
• 当p长度为1的时候要单独处理，因为这时候我们不用判断p的第1位是否是‘*’了。处理这一段代码如下：

1. if(p.length() == 1){
           if(p.charAt(0) == ‘.‘ && s.length() == 1)
               return true;
           return s.equals(p);
   }

• 其他情况就要按照p的第1位是否为‘*‘来分了。如果不为’*‘，那么p和s的第0位必须匹配（相等或p第0位为‘.‘），否则p和s不匹配，这样的例子类似的有s = "ab", p = "c*b"。如果为‘*‘，我们就按照匹配0位，1位，2位.....的方式递归试探，类似的例子有s = "abbc", p = "ab*bbc",此时‘*‘并不匹配s中的任何字符，再有s = "aa",p = "a*",此时‘*‘匹配s中的两个a。

代码如下：

 1 public class Solution {
 2     public boolean isMatch(String s, String p) {
 3          if(p.length() == 0)
 4             return s.length() == 0;
 5
 6         if(s.length() == 0){
 7             if(p.length() == 1)
 8                 return false;
 9             if(p.charAt(1) == ‘*‘)
10                 return isMatch(s, p.substring(2));
11             return false;
12         }
13
14
15         if(p.length() == 1){
16             if(p.charAt(0) == ‘.‘ && s.length() == 1)
17                 return true;
18             return s.equals(p);
19         }
20
21         if(p.length() >= 2 && p.charAt(1) != ‘*‘){
22             //if p(1) is not *, we need p(0) equals to s(0) or p(0) equals ‘.‘
23             if(p.charAt(0) == s.charAt(0) || p.charAt(0) == ‘.‘ && s.length() != 0)
24                 //check if the left is also match
25                 return isMatch(s.substring(1), p.substring(1));
26             return false;
27         }
28         else{
29             //if p(1) is ‘*‘,we check how many we can match from this * by trying
30             int i = 0;
31             char now = p.charAt(0);
32             while(i<s.length() && (s.charAt(i) == now || now == ‘.‘)){
33                 if(isMatch(s.substring(i+1),p.substring(2)))
34                         return true;
35                 i++;
36             }
37             //this means we don‘t use this ‘*‘ to match any character, just skip it
38             return isMatch(s, p.substring(2));
39         }
40     }
41 }

【leetcode刷题笔记】Regular Expression Matching