题目

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") → falseisMatch("aa","aa") → trueisMatch("aaa","aa") → falseisMatch("aa", "a*") → trueisMatch("aa", ".*") → trueisMatch("ab", ".*") → trueisMatch("aab", "c*a*b") → true

方法

使用递归的思想,分为两种情况：

1. p[i + 1] != '*', 返回 p[i] == s[j] && (s(j + 1), p(i + 1)):递归求解

2. p[i + 1] == '*',分p[i] != s[j], 递归求解（s(j), p(i + 2)）

     p[i] == s[j] 递归求解（s(j), p(i + 2)）以及 (s(j + 1) ,p (i + 2))

private boolean getMatch(String s, int lenS, int curS, String p, int lenP, int curP) {				if (curP == lenP) {			return curS == lenS;		}		if (curP + 1 == lenP || p.charAt(curP + 1) != '*') {			if (curS == lenS) {				return false;			}			return (p.charAt(curP) == s.charAt(curS) || p.charAt(curP) == '.') && getMatch(s, lenS, curS + 1, p, lenP, curP + 1);		}				while (curS < lenS && (s.charAt(curS) == p.charAt(curP) || p.charAt(curP) == '.')) {			if (getMatch(s, lenS, curS, p, lenP, curP + 2)) {				return true;			}			curS++;		}		return getMatch(s, lenS, curS, p, lenP, curP + 2);	}    public boolean isMatch(String s, String p) {    	if ((s == null && p == null) || (s.length() == 0 && p.length() == 0)) {    		return true;    	}    	int lenS = s.length();    	int lenP = p.length();        return getMatch(s, lenS, 0, p, lenP, 0);    }