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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：
正则表达式的匹配，‘.‘能匹配任何一个字符，‘*‘之前必须有一个字符，两个结合起来表示之前那个字符出现0到无穷次。
解法：
一定要注意‘*‘必须结合前面的字符一起使用。
本文及代码引用自：http://harrifeng.github.io/algo/leetcode/regular-expression-matching.html
代码如下
public static boolean isMatch(String s, String p) { if (p.length() == 0) return s.length() == 0; // length == 1 is the case that is easy to forget. // as p is subtracted 2 each time, so if original // p is odd, then finally it will face the length 1 if (p.length() == 1) return (s.length() == 1) && (p.charAt(0) == s.charAt(0) || p.charAt(0) == ‘.‘); // next char is not ‘*‘: must match current character if (p.charAt(1) != ‘*‘) { if (s.length() == 0) return false; else return (s.charAt(0) == p.charAt(0) || p.charAt(0) == ‘.‘) && isMatch(s.substring(1), p.substring(1)); }else{ // next char is * while (s.length() > 0 && (p.charAt(0) == s.charAt(0) || p.charAt(0) == ‘.‘)) { if (isMatch(s, p.substring(2))) return true; s = s.substring(1); } return isMatch(s, p.substring(2)); } }
reference： http://www.cnblogs.com/springfor/p/3893593.html
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原文地址：http://www.cnblogs.com/hygeia/p/5469756.html
