Regular Expression Matching
2014-12-02 14:11
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Implement regular expression matching with support for
基本思路就是先看字符串s和p的从i和j开始的子串是否匹配,用递归的方法直到串的最后,最后回溯回来得到结果。假设现在走到s的i位置,p的j位置,情况分为下列两种:
(1)p[j+1]不是'*'。情况比较简单,只要判断当前s的i和p的j上的字符是否一样(如果有p在j上的字符是'.',也是相同),如果不同,返回false,否则,递归下一层i+1,j+1;
(2)p[j+1]是'*'。那么此时看从s[i]开始的子串,假设s[i],s[i+1],...s[i+k]都等于p[j]那么意味着这些都有可能是合适的匹配,那么递归对于剩下的(i,j+2),(i+1,j+2),...,(i+k,j+2)都要尝试(j+2是因为跳过当前和下一个'*'字符)。
C++实现代码如下:
'.'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
基本思路就是先看字符串s和p的从i和j开始的子串是否匹配,用递归的方法直到串的最后,最后回溯回来得到结果。假设现在走到s的i位置,p的j位置,情况分为下列两种:
(1)p[j+1]不是'*'。情况比较简单,只要判断当前s的i和p的j上的字符是否一样(如果有p在j上的字符是'.',也是相同),如果不同,返回false,否则,递归下一层i+1,j+1;
(2)p[j+1]是'*'。那么此时看从s[i]开始的子串,假设s[i],s[i+1],...s[i+k]都等于p[j]那么意味着这些都有可能是合适的匹配,那么递归对于剩下的(i,j+2),(i+1,j+2),...,(i+k,j+2)都要尝试(j+2是因为跳过当前和下一个'*'字符)。
C++实现代码如下:
#include<iostream> #include<string> using namespace std; class Solution { public: bool isMatch(const char *s, const char *p) { return helper(string(s),string(p),0,0); } bool helper(const string s,const string p,size_t i,size_t j) { if(j==p.length()) return i==s.length(); //p.charAt(j+1)!='*' if(j==p.length()-1||p[j+1]!='*') { if(i==s.length()||(s[i]!=p[j]&&p[j]!='.')) return false; else return helper(s,p,i+1,j+1); } //p.charAt(j+1)=='*' while(i<s.length()&&(s[i]==p[j]||p[j]=='.')) { if(helper(s,p,i,j+2)) return true; i++; } return helper(s,p,i,j+2); } }; int main() { Solution ss; const char *s="aa"; const char *p=".*bbb"; cout<<ss.isMatch(s,p)<<endl; }
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