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);  }};