leetcode 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>

class Solution {
 public:
  bool isMatch(const char *s, const char *p) {
    int slen = strlen(s), plen = strlen(p), i, j;
    bool dp[500][500];
    memset(dp,false, sizeof(dp));
    dp[0][0] = true;
    for (i = 1; i <= plen; ++i) {
      if (p[i] == '*') {
        dp[i][0] = dp[i + 1][0] = dp[i - 1][0];
        for (j = 1; j <= slen; ++j)
          dp[i][j] = dp[i + 1][j] = (dp[i][j - 1] && (p[i - 1] == s[j - 1] || p[i - 1] == '.') || dp[i - 1][j]);
        ++i;        
      }
      else 
        for (j = 1; j <= slen; ++j)
          dp[i][j] = dp[i - 1][j - 1] && (p[i - 1] == s[j - 1] || p[i - 1] == '.');
    }
    return dp[plen][slen];
  }
};

OR

class Solution {
 public:
  bool isMatch(const char *s, const char *p) {
    const char *ss = s, *sbegin = s, *pbegin = p, *pp = p;
    bool dp[500][500];
    memset(dp, false, sizeof(dp));
    int slen = strlen(s), plen = strlen(p), i, j;
    dp[0][0] = true;
    for (i = 1; i <= plen; ++i) {
      if (p[i - 1] == '*') 
        dp[i][0] = dp[i - 1][0] = dp[i - 2][0];
      
      for (j = 1; j <= slen; ++j) {
        if (p[i - 1] == '*') 
          dp[i][j] = dp[i - 1][j] = ((dp[i - 2][j - 1] || dp[i][j - 1]) && (p[i - 2] == s[j - 1] || p[i - 2] == '.')) || dp[i - 2][j]; 
        else 
          dp[i][j] = dp[i - 1][j - 1] && (p[i - 1] == s[j - 1] || p[i - 1] == '.');
      }
    }
    return dp[plen][slen];
  }
};

Take notice that 

dp[i - 2][j - 1] || dp[i][j - 1]

Can't be written as 

dp[i - 2][j - 1]

For exmaple, "aa" and "a*"

recursive:

class Solution {
 public:
  bool isMatch(const char *s, const char *p) {
    if (*p == '\0')
      return *s == '\0';
    if (*(p + 1) == '*') {
      for (const char *str = s; (*str == *p || *p == '.' && *str); ++str) 
        if(isMatch(str + 1, p + 2))
          return true;
      return isMatch(s, p + 2);
    }  
    else
      return (*p == *s || *p == '.' && *s) && isMatch(s + 1, p + 1);
  }
};

Give a wrong code, find out the differences:

class Solution {
 public:
  bool isMatch(const char *s, const char *p) {
    if (*s == *p && *s == '\0')
      return true;
    if (*(p + 1) == '*') {
      for (const char *str = s; *str == *p || *p == '.' || *str == *s; ++str)
        if (isMatch(str, p + 2))
          return true;
      return false;
    }
    else
      return (*p == *s || *p == '.') && isMatch(s + 1, p + 1);
  }
};