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) != '\0') {

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 == '\0') return (*s == '\0');

 5         // match single '\0', '.', 'a'...

 6         if (*(p + 1) != '*') { 

 7             return ((*s == *p || (*p == '.' && *s != '\0')) && isMatch(s + 1, p + 1));

 8         }

 9         

10         // match a*, .*

11         while ((*s == *p || (*p == '.' && *s != '\0'))) {

12             if (isMatch(s++, p + 2)) return true;

13         }

14         

15         // ignore a*， *p != '.' && *s != *p

16         return isMatch(s, p + 2);

17     }

18 };