string-10.Regular Expression Matching
2017-12-23 23:04
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题目描述:
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
题目解析:
1. 此题为正则式匹配。‘.’可以代表任意一个字符,而‘*’代表它前面的字符可以出现n次(n为非负整数)。匹配成功返回true,匹配失败返回false。2. 理解此题最为简单易懂就是利用递归。废话不多说,直接上代码,注释我写的很清楚,个人理解此段代码时间复杂度O(n)。
代码如下:
class Solution { public: bool isMach(const char *s, const char *p) { if(*s=='\0' && *p == '\0') return true; else if(*s != '\0' && *p == '\0') return false; // 如果*(p+1)是*, if(*(p+1) == '*') { if(*p == *s || *p == '.' && *s != '\0') // 有两种情况,依次分别是:*代表和0个字符匹配,不起作用;*代表的已经和当前S匹配了。继续看S的下一个,再看*到底可以代表几个连续匹配 return isMach(s, p+2) || isMach(s+1, p); return isMach(s, p+2); } // 如果s和p当前指向的值是一样的; else if(*s == *p || *p == '.' && *s != '\0') return isMach(s+1, p+1); else return false; } bool isMatch(string s, string p) { if(s.empty() && p.empt 4000 y()) return true; return isMach(s.c_str(), p.c_str()); } };
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