LeetCode #10 Regular Expression Matching
2017-03-19 22:33
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Description
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
Analysis
题目难度为:Hard本题确实很难,让我思考了很久的时间,但是实现的代码却很简洁,下面对题目进行分析。题目中默认了输入的都是正确的正则表达式,不考虑错误处理。我们每次检测两个字符,根据第二个字符是什么来进行判断。情况1,第二个字符为,此时可以直到,正则表达式能够匹配0个或者多个第一个字符,但是我们不知道具体需要匹配多少个,那么可以用递归的思想,让计算机帮我们思考——匹配0个或者匹配1个(匹配一个的情况就包含了匹配多个的情况);情况2,第二个字符不是那么,只要匹配第一个字符就可以了。
Code(c++)
class Solution { public: bool isMatch(string s, string p) { if (p.empty()) return s.empty(); if ('*' == p[1]) return (isMatch(s, p.substr(2)) || !s.empty() && (s[0] == p[0] || '.' == p[0]) && isMatch(s.substr(1), p)); else return !s.empty() && (s[0] == p[0] || '.' == p[0]) && isMatch(s.substr(1), p.substr(1)); } };
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