LeetCode——010

/* 题目： 10. Regular Expression Matching My Submissions QuestionEditorial SolutionTotal Accepted: 77160 Total Submissions: 352286 Difficulty: HardImplement 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") → falseisMatch("aa","aa") → trueisMatch("aaa","aa") → falseisMatch("aa", "a*") → trueisMatch("aa", ".*") → trueisMatch("ab", ".*") → true*//* 解题思路：需要用递归Recursion来解，大概思路如下：- 若p为空，若s也为空，返回true，反之返回false- 若p的长度为1，若s长度也为1，且相同或是p为'.'则返回true，反之返回false- 若p的第二个字符不为*，若此时s为空返回false，否则判断首字符是否匹配，且从各自的第二个字符开始调用递归函数匹配- 若p的第二个字符为*，若s不为空且字符匹配，调用递归函数匹配s和去掉前两个字符的p，若匹配返回true，否则s去掉首字母- 返回调用递归函数匹配s和去掉前两个字符的p的结果*/代码如下：class Solution {public:    bool isMatch(string s, string p) {         if(p.empty())return s.empty();         if(p.size()==1) {           return (s.size()==1 && (s[0]==p[0]||p[0]=='.'));            }         if(p[1]!='*'){             if(s.empty())return false;             return (s[0]==p[0]||p[0]=='.')&&isMatch(s.substr(1),p.substr(1));         }         while(!s.empty()&&(s[0]==p[0]||p[0]=='.')){             if(isMatch(s,p.substr(2)))return true;             s=s.substr(1);         }         return isMatch(s,p.substr(2));    }};