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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
法I：回溯法。在下一个字符是*时 要进行backtrack。
class Solution { public: bool isMatch(string s, string p) { return dfsIsMatch(s,p,0,0); } bool dfsIsMatch(const string& s, const string& p, int sIndex, int pIndex){ if (p[pIndex] == ‘\0‘) //结束条件：s若是‘\0‘，p必须也是‘\0‘ return s[sIndex] == ‘\0‘; if (p[pIndex+1] == ‘*‘) { /* ‘.‘ means any character (except ‘\0‘) * ‘.‘ means repeat 0 or more times * ‘.*‘ means repeat ‘.‘ 0 or more times */ while ((s[sIndex] != ‘\0‘ && p[pIndex] == ‘.‘) || s[sIndex] == p[pIndex]) { //‘.‘可以与除‘\0‘以外的任何字符匹配 if (dfsIsMatch(s, p, sIndex, pIndex+2)) //p[pIndex] repeat 0 times return true; sIndex += 1;//p[pIndex]在原基础上repeat次数+1 } return dfsIsMatch(s, p, sIndex, pIndex+2); //when s[sIndex] != p[pIndex] && p[pIndex] != ‘.‘(此时只有s[sIndex]==p[pIndex]==‘\0‘时可能return true) } else if ((s[sIndex] != ‘\0‘ && p[pIndex] == ‘.‘) || s[sIndex] == p[pIndex]) { return dfsIsMatch(s, p, sIndex + 1, pIndex + 1); } return false; } };
法II：动态规划。设二位数组dp[i][j]来表示两个string的状态关系，dp[i][j]=true，表示s[0..i]匹配p[0..j]的结果。
10. Regular Expression Matching
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原文地址：http://www.cnblogs.com/qionglouyuyu/p/5860126.html
