10. 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

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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.

class Solution { public: bool isMatch(string s, string p) { int m = s.size(), n = p.size(); vector<vector<bool>> dp(m + 1, vector<bool>(n + 1,false)); dp[0][0] = true; for (int i = 0; i <= m; i++){ for (int j = 1; j <= n; j++){ if (p[j - 1] == '*'){ dp[i][j] = dp[i][j - 2] || (i > 0 && dp[i - 1][j] && (s[i - 1] == p[j - 2] || p[j - 2] == '.')); } else{ if (i > 0 && (s[i-1] == p[j-1] || p[j-1] == '.')){ dp[i][j] = dp[i - 1][j - 1]; } } } } return dp[m][n]; } };

44. Wildcard Matching

Implement wildcard pattern matching with support for '?' and '*'.

'?' Matches any single character. '*' Matches any sequence of characters (including the empty sequence). 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", "*") → true isMatch("aa", "a*") → true isMatch("ab", "?*") → true isMatch("aab", "c*a*b") → false

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这个跟上面的区别是，这里*是通配符可以匹配任何字符，而不是它前面的那个字符。

class Solution { public: bool isMatch(string s, string p) { int m = s.size(), n = p.size(); vector<vector<bool>> dp(m + 1, vector<bool>(n + 1, false)); dp[0][0] = true; for (int i = 0; i <= m; i++){ for (int j = 1; j <= n; j++){ if (p[j - 1] == '*'){ dp[i][j] = dp[i][j - 1] || (i > 0 && dp[i - 1][j]);// } else{ if (i > 0 && (s[i - 1] == p[j - 1] || p[j - 1] == '?')){ dp[i][j] = dp[i - 1][j - 1]; } } } } return dp[m][n]; } };