LeetCode | Regular Expression Matching

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

思路是分别处理a*、.*、.、a的情况。

三周前写的lengthy code如下：

 1 class Solution {
 2 public:
 3     bool isMatch(const char *s, const char *p) {
 4         int sn = strlen(s);
 5         int pn = strlen(p);
 6         return recursive(s, sn, p, pn);
 7     }
 8     
 9     bool recursive(const char* s, int sn, const char* p, int pn) {
10         if (sn == 0 && pn == 0) return true;
11         if (pn == 0) return false;
12         
13         if (*(p + 1) != '') {
14             if (*(p + 1) == '*') {
15                 if (*p == '.') { // .*
16                     int n = 0;
17                     while (n <= sn) {
18                         if (recursive(s + n, sn - n, p + 2, pn - 2)) return true;
19                         n++;
20                     }
21                 } else { // a*
22                     int n = 0;
23                     while (n <= sn && *(s + n) == *p) {
24                         if (recursive(s + n, sn - n, p + 2, pn - 2)) return true;
25                         n++;
26                     }
27                     if (recursive(s + n, sn - n, p + 2, pn - 2)) return true;
28                 }
29             } else {
30                 if (*p != '.' && *s != *p) return false;
31                 if (recursive(s + 1, sn - 1, p + 1, pn - 1)) return true;
32             }
33         } else {
34             if (*p != '.' && *s != *p) return false;
35             if (recursive(s + 1, sn - 1, p + 1, pn - 1)) return true;
36         }
37     }
38 };

今天看了Leetcode上1337的代码真是羞愧啊。http://leetcode.com/2011/09/regular-expression-matching.html

重写了一遍。思路还是一样。

 1 class Solution {
 2 public:
 3     bool isMatch(const char *s, const char *p) {
 4         if (*p == '') return (*s == '');
 5         // match single '', '.', 'a'...
 6         if (*(p + 1) != '*') { 
 7             return ((*s == *p || (*p == '.' && *s != '')) && isMatch(s + 1, p + 1));
 8         }
 9         
10         // match a*, .*
11         while ((*s == *p || (*p == '.' && *s != ''))) {
12             if (isMatch(s++, p + 2)) return true;
13         }
14         
15         // ignore a*， *p != '.' && *s != *p
16         return isMatch(s, p + 2);
17     }
18 };