LeetCode – Regular Expression Matching in Java

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本文探讨了支持'.'和'*'的正则表达式匹配问题，并提供了两种Java解决方案。第一种方案简洁明了，第二种则更加易于理解，通过递归方式解决了不同情况下的匹配问题。

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") return false
isMatch("aa","aa") return true
isMatch("aaa","aa") return false
isMatch("aa", "a*") return true
isMatch("aa", ".*") return true
isMatch("ab", ".*") return true
isMatch("aab", "c*a*b") return true

1. Analysis

First of all, this is one of the most difficulty problems. It is hard to think through all different cases. The problem should be simplified to handle 2 basic cases:

the second char of pattern is "*"
the second char of pattern is not "*"

For the 1st case, if the first char of pattern is not ".", the first char of pattern and string should be the same. Then continue to match the remaining part.

For the 2nd case, if the first char of pattern is "." or first char of pattern == the first i char of string, continue to match the remaining part.

2. Java Solution 1 (Short)

The following Java solution is accepted.

public class Solution {
    public boolean isMatch(String s, String p) {
 
        if(p.length() == 0)
            return s.length() == 0;
 
        //p's length 1 is special case    
        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 len = s.length();
 
            int i = -1; 
            while(i<len && (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;
        } 
    }
}

3. Java Solution 2 (More Readable)

public boolean isMatch(String s, String p) {
	// base case
	if (p.length() == 0) {
		return s.length() == 0;
	}
 
	// special case
	if (p.length() == 1) {
 
		// if the length of s is 0, return false
		if (s.length() < 1) {
			return false;
		}
 
		//if the first does not match, return false
		else if ((p.charAt(0) != s.charAt(0)) && (p.charAt(0) != '.')) {
			return false;
		}
 
		// otherwise, compare the rest of the string of s and p.
		else {
			return isMatch(s.substring(1), p.substring(1));
		}
	}
 
	// case 1: when the second char of p is not '*'
	if (p.charAt(1) != '*') {
		if (s.length() < 1) {
			return false;
		}
		if ((p.charAt(0) != s.charAt(0)) && (p.charAt(0) != '.')) {
			return false;
		} else {
			return isMatch(s.substring(1), p.substring(1));
		}
	}
 
	// case 2: when the second char of p is '*', complex case.
	else {
		//case 2.1: a char & '*' can stand for 0 element
		if (isMatch(s, p.substring(2))) {
			return true;
		}
 
		//case 2.2: a char & '*' can stand for 1 or more preceding element, 
		//so try every sub string
		int i = 0;
		while (i<s.length() && (s.charAt(i)==p.charAt(0) || p.charAt(0)=='.')){
			if (isMatch(s.substring(i + 1), p.substring(2))) {
				return true;
			}
			i++;
		}
		return false;
	}
}