LeetCode.010 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

题意
带通配符带字符串匹配。

.        匹配任意的单个字符

*        匹配前一个字符的零个或者多个出现

思路

借鉴代码之美中的解决方案，利用递归的方法，代码简洁，易懂。

//摘自《代码之美》
// 字符     含义
// .        匹配任意的单个字符
// ^        匹配输入字符串的开头
// $        匹配输入字符串的结尾
// *        匹配前一个字符的零个或者多个出现
#include <stdio.h>
int matchhere(char *regexp, char *text);

int matchstar(int c, char *regexp, char *text) {// matchstar: search for c*regexp at beginning of text
   do {// a * matches zero or more instances
       if (matchhere(regexp, text)) return 1;
   } while (*text != '\0' && (*text++ == c || c == '.'));
   return 0;
}
int matchhere(char *regexp, char *text) {// matchhere: search for regexp at beginning of text
   if (regexp[0] == '\0') return 1;
   if (regexp[1] == '*') return matchstar(regexp[0], regexp+2, text);
   if (regexp[0] == '$' && regexp[1] == '\0') return *text == '\0';
   if (*text!='\0' && (regexp[0]=='.' || regexp[0]==*text)) return matchhere(regexp+1, text+1);
   return 0;
}

int match(char *regexp, char *text) {// match: search for regexp anywhere in text
    if (regexp[0] == '^') return matchhere(regexp+1, text);
    do {// must look even if string is empty
        if (matchhere(regexp, text)) return 1;
    } while (*text++ != '\0');
    return 0;
}

下面是自己的关于此题的代码。

bool isMatch(char* s, char* p);
bool matchstar(int c,char *s,char *p)  
{
    do {// a * matches zero or more instances
        if (isMatch(s, p)) return 1;
    } while (*s != '\0' && (*s++ == c || c == '.'));
    return 0;
}
bool isMatch(char* s, char* p) {
    if(p[0]=='\0' && s[0]=='\0') return 1;
    if(p[1]=='*') return matchstar(p[0], s,p+2);
    if(*s!='\0' && (p[0]=='.' || *s==p[0])) return isMatch(s+1,p+1);
    return 0;
    
}