0010.Regular Expression Matching(H)
jjc
. Regular Expression Matching(hard) Given an input string (s) and a pattern (p), 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). Note: s could be empty and contains only lowercase letters a-z. p could be empty and contains only lowercase letters a-z, and characters like . or *. Example : Input: s = "aa" p = "a" Output: false Explanation: "a" does not match the entire string "aa". Example : Input: s = "aa" p = "a*" Output: true Explanation: '*' means zero or more of the precedeng element, 'a'. Therefore, by repeating 'a' once, it becomes "aa". Example : Input: s = "ab" p = ".*" Output: true Explanation: ".*" means "zero or more (*) of any character (.)". Example : Input: s = "aab" p = "c*a*b" Output: true Explanation: c can be repeated times, a can be repeated time. Therefore it matches "aab". Example : Input: s = "mississippi" p = "mis*is*p*." Output: false Accepted ,,,
class Solution_S4ms { public: bool isMatch(string s, string p) { ](); ; i < s.length() + ; ++i) { dp[i] = ]; memset(dp[i], , p.length()+); } dp[s.size()][p.size()] = true; ; i--){ ; j >= ; j--) { bool first_match = (i < s.length() && (p[j] == s[i] || p[j] == '.')); < p.length() && p[j+] == '*') { dp[i][j] = dp[i][j+] || first_match && dp[i+][j]; } else { dp[i][j] = first_match && dp[i+][j+]; } } } ][]; } }; class Solution_S8ms { public: bool isMatch(string s, string p) { int m = s.size(), n = p.size(); vector<vector<, vector<, false)); dp[][] = true; ; i <= m; ++i) { ; j <= n; ++j) { && p[j - ] == '*') { dp[i][j] = dp[i][j - ] || (i > && (s[i - ] == p[j - ] || p[j - ] == ][j]); } else { dp[i][j] = i > && dp[i - ][j - ] && (s[i - ] == p[j - ] || p[j - ] == '.'); } } } return dp[m][n]; } };
/* https://www.cnblogs.com/grandyang/p/4461713.html * dp[i][j]表示s[0,i)和p[0,j)是否match 1. P[i][j] = P[i - 1][j - 1], if p[j - 1] != '*' && (s[i - 1] == p[j - 1] || p[j - 1] == '.'); 2. P[i][j] = P[i][j - 2], if p[j - 1] == '*' and the pattern repeats for 0 times; 3. 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_grandyang { public: bool isMatch(string s, string p) { int m = s.size(), n = p.size(); vector<vector<, vector<, false)); dp[][] = true; ; i <= m; ++i) { ; j <= n; ++j) { && p[j - ] == '*') { dp[i][j] = dp[i][j - ] || (i > && (s[i - ] == p[j - ] || p[j - ] == ][j]); } else { dp[i][j] = i > && dp[i - ][j - ] && (s[i - ] == p[j - ] || p[j - ] == '.'); } } } return dp[m][n]; } };
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