regular expression matching 正则匹配

/*
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

*
The asterisk indicates zero or more occurrences of the preceding element.
For example, ab*c matches "ac", "abc", "abbc", "abbbc", and so on.
“*”代表之前的那个字符可以是0个或者任意多个
*/
#include <stdio.h>
#include <string.h>

/*
采用递归解决：
若p为空，s也为空返回true，否则返回false
若p[1]=='*'时，’*‘代表0个前面的字符时，把p的前两个丢掉进行递归；’*‘代表多个前一个字符
时，若果s字符长度大于1，则比较完s[0]==p[0]后，把s的第一个字符丢掉然后进行递归。
若p[1]!='*'时，且s_len不为零，则把s p的第一个字符都丢掉并递归。
*/
int isMatch(char* s, char* p){
int s_len=strlen(s);
int p_len=strlen(p);
if(p_len==0) return s_len==0;
if(p_len>1 && p[1]=='*')
return isMatch(s,&(p[2])) || (s_len!=0 && (s[0]==p[0]||p[0]=='.') && isMatch(&(s[1]),p));
else
return s_len!=0 && (s[0]==p[0] || p[0]=='.') && isMatch(&(s[1]),&(p[1]));
}
/*
参考：https://discuss.leetcode.com/topic/17852/9-lines-16ms-c-dp-solutions-with-explanations
采用动态规划：
This problem has a typical solution using Dynamic Programming. We define the state P[i][j] to be true
if s[0..i) matches p[0..j) and false otherwise. Then the state equations are:
P[i][j] = P[i-1][j-1],if p[j-1] != '*' && (s[i-1] == p[j-1] || p[j-1] == '.');
P[i][j] = P[i][j-2], if p[j-1] == '*' and the pattern repeats for 0 times;
P[i][j] = P[i-1][j] && (s[i-1] == p[j-2] || p[j-2] == '.'), if p[j-1] == '*' and the pattern repeats
for at least 1 times.
*/
int isMatch_dp(char *s,char *p)
{
int s_len=strlen(s);
int p_len=strlen(p);
int dp[s_len+1][p_len+1];
memset(dp,0,sizeof(dp));
dp[0][0]=1;//s p 长度为0时 为真
int i,j;
for(i=0;i<=s_len;i++)
for(j=1;j<=p_len;j++)
{
if(j>1 && p[j-1]=='*')
dp[i][j]=dp[i][j-2] || (i>0 && (s[i-1]==p[j-2] || p[j-2]=='.') && dp[i-1][j]);
else
dp[i][j]=i>0 && dp[i-1][j-1] && (s[i-1]==p[j-1] || p[j-1]=='.');

}
return dp[s_len][p_len];
}
int main(int argc,char **argv)
{
char s[]="aa";
char p[]=".*";
printf("%d\n",isMatch(s,p));
printf("%d\n",isMatch_dp(s,p));
return 0;
}