Wildcard Matching

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/*from: http://www.jiuzhang.com/solutions/wildcard-matching/动态规划法dp[i][j]表示s的前i个字符与p的前j个字符匹配class Solution {public:    bool isMatch(const char *s, const char *p) {        if (s == NULL || p == NULL) {            return false;        }        int n = strlen(s);        int m = strlen(p);        int f[n + 1][m + 1];        memset(f, false, sizeof(f));        f[0][0] = true;        for (int i = 1; i <= n; i++)            f[i][0] = false;        for (int i = 1; i <= m; i++)            f[0][i] = f[0][i - 1] && p[i - 1] == '*';        for (int i = 1; i <= n; i++) {            for (int j = 1; j <= m; j++) {                if (p[j - 1] == '*') {                    f[i][j] = f[i - 1][j] || f[i][j - 1];                } else if (p[j - 1] == '?') {                    f[i][j] = f[i - 1][j - 1];                } else {                    f[i][j] = f[i - 1][j - 1] && (s[i - 1] == p[j - 1]);                }            }        } // for        return f[n][m];    }};*/class Solution {public:    bool isMatch(string s, string p)     {        int sLen = s.length();        int pLen = p.length();        if( sLen <= 0 || pLen <= 0 )        {            return false;        }        bool dp[ sLen + 1 ][ pLen + 1 ];        memset( dp, false, sizeof(dp) );        dp[0][0] = true;        for( int i = 1; i <= sLen; i++ )        {            dp[i][0] =  false;        }        for( int i = 1; i <= pLen; i++ )        {            dp[0][i] = dp[0][ i - 1 ] && ( '*' == p[ i - 1 ] );        }        for( int i = 1; i <= sLen; i++ )        {            for( int j = 1; j <= pLen; j++ )            {                if( '*' == p[ j - 1 ] )                {                    dp[i][j] = dp[i][ j - 1 ] || dp[ i - 1 ][j];                }                else if( '?' == p[ j - 1 ] )                {                    dp[i][j] = dp[ i - 1 ][ j - 1 ];                }                else                {                    dp[i][j] = dp[ i - 1 ][ j - 1 ] && ( s[ i - 1 ] == p[ j - 1 ] );                }            }        }        return dp[sLen][pLen];    }};//////////////////////////////////////////////////////////////////////////////////////*from: http://www.jiuzhang.com/solutions/wildcard-matching/动态规划法dp[i][j]表示s的前i个字符与p的前j个字符匹配class Solution {public:    bool isMatch(const char *s, const char *p) {        if (s == NULL || p == NULL) {            return false;        }        int n = strlen(s);        int m = strlen(p);        int f[n + 1][m + 1];        memset(f, false, sizeof(f));        f[0][0] = true;        for (int i = 1; i <= n; i++)            f[i][0] = false;        for (int i = 1; i <= m; i++)            f[0][i] = f[0][i - 1] && p[i - 1] == '*';        for (int i = 1; i <= n; i++) {            for (int j = 1; j <= m; j++) {                if (p[j - 1] == '*') {                    f[i][j] = f[i - 1][j] || f[i][j - 1];                } else if (p[j - 1] == '?') {                    f[i][j] = f[i - 1][j - 1];                } else {                    f[i][j] = f[i - 1][j - 1] && (s[i - 1] == p[j - 1]);                }            }        } // for        return f[n][m];    }};*/class Solution {public:    bool myMatch(const char *s, const char *p) {        if (s == NULL || p == NULL) {            return false;        }        int n = strlen(s);        int m = strlen(p);        vector< vector<bool> > f( n + 1, vector<bool>( m + 1, false ) );        //memset(f, false, sizeof(f));        f[0][0] = true;        for (int i = 1; i <= n; i++)            f[i][0] = false;        for (int i = 1; i <= m; i++)            f[0][i] = f[0][i - 1] && p[i - 1] == '*';        for (int i = 1; i <= n; i++) {            for (int j = 1; j <= m; j++) {                if (p[j - 1] == '*') {                    f[i][j] = f[i - 1][j] || f[i][j - 1];                } else if (p[j - 1] == '?') {                    f[i][j] = f[i - 1][j - 1];                } else {                    f[i][j] = f[i - 1][j - 1] && (s[i - 1] == p[j - 1]);                }            }        } // for        return f[n][m];    }    bool isMatch(string s, string p)     {        return myMatch( s.c_str(), p.c_str() );    }};
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