LeetCode——Regular Expression Matching

标签：leetcode

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

‘ . ‘ 匹配任何单字符。

‘ * ‘匹配0或多个前向元素。

使用递归进行判断。总体上可以分成两种情况，一种是以 ‘ * ‘开头的，另一种不是。

public class RegularExpressionMatching {
	public static void main(String[] args) {
		System.out.println(isMatch("aa","a"));
		System.out.println(isMatch("aa","aa"));
		System.out.println(isMatch("aaa","aa"));
		System.out.println(isMatch("aa", "a*"));
		System.out.println(isMatch("aa", ".*"));
		System.out.println(isMatch("ab", ".*"));
		System.out.println(isMatch("aab", "c*a*b"));
	}
	public static boolean isMatch(String s,String p){
		if(p.length() == 0)
			return s.length() == 0;
		if(p.length() == 1 || p.charAt(1) != '*'){
			if(s.length()  < 1 || (p.charAt(0) != '.' && s.charAt(0) != p.charAt(0)))
				return false;
			return isMatch(s.substring(1),p.substring(1));
		}else{
			int i = -1;
			while(i < s.length() && (i < 0 || p.charAt(0) == '.' || p.charAt(0) == s.charAt(i))){
				if(isMatch(s.substring(i+1),p.substring(2)))
					return true;
				i++;
			}
			return false;
		}
		
	}
}

Reference：http://www.programcreek.com/2012/12/leetcode-regular-expression-matching-in-java/

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LeetCode——Regular Expression Matching

标签：leetcode

原文地址：http://blog.csdn.net/laozhaokun/article/details/37351707