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题目：（ DP ,BackTracking, String）
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
题解：
本文及代码引用自：http://harrifeng.github.io/algo/leetcode/regular-expression-matching.html
public class Solution { public boolean isMatch(String s, String p) { if(p.length()==0) return s.length()==0; if(p.length()==1) return (s.length()==1)&&(p.charAt(0)==s.charAt(0)||p.charAt(0)==‘.‘); if(p.charAt(1)!=‘*‘) { if(s.length()==0) return false; else return isMatch(s.substring(1),p.substring(1))&&(s.charAt(0)==p.charAt(0)||p.charAt(0)==‘.‘); } else { while(s.length()>0&&(s.charAt(0)==p.charAt(0)||p.charAt(0)==‘.‘)) { if(isMatch(s,p.substring(2))) return true; s=s.substring(1); } return isMatch(s,p.substring(2)); } } }
[leetcode] Regular Expression Matching
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原文地址：http://www.cnblogs.com/fengmangZoo/p/4183804.html
