[leetcode] Regular Expression Matching

最新推荐文章于 2021-01-27 05:31:01 发布

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本文介绍了一种正则表达式匹配的实现方法，支持‘.’和‘*’两种特殊字符，通过递归的方式实现了字符串与正则表达式的完全匹配。提供了详细的Java与C++代码示例，并解释了每种情况下的匹配逻辑。

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from : https://leetcode.com/problems/regular-expression-matching/

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

public class Solution {
    public boolean isMatch(String s, String p) {
		if (null == s || null == p) {
			return s == p;
		}

		int lenS = s.length();
		int lenP = p.length();

		if (0 == lenP) {
			return 0 == lenS;
		}

		if (1 == lenP) {
			if (p.equals(s) || ".".equals(p) && 1 == s.length()) {
				return true;
			} else {
				return false;
			}
		}
		
		if ('*' != p.charAt(1)) {
			if (0 < s.length() && (p.charAt(0) == s.charAt(0) || '.' == p.charAt(0))) {
				return isMatch(s.substring(1), p.substring(1));
			}
			return false;
		} else {
			while (0 < s.length() && (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));
		}

	}
}

class Solution {
public:
    bool isMatch(string s, string p) {
        int ls = s.size(), lp = p.size();
        if(0 == lp) return 0 == ls;
        if(1 == lp) {
            return 1 == ls && ('.' == p[0] || p[0] == s[0]);
        }
        
        if('*' != p[1]) {
            if(0 < ls && (s[0] == p[0] || '.' == p[0])) {
                return isMatch(s.substr(1), p.substr(1));
            }
            return false;
        } else {
            while(0 < s.size() && (s[0] == p[0] || '.' == p[0])) {
                if(isMatch(s, p.substr(2))) {
                    return true;
                }
                s = s.substr(1);
            }
            return isMatch(s, p.substr(2));
        }
    }
};