LeetCode Regular Expression Matching_最小三元组距离leetcode-优快云博客

本文介绍了一种正则表达式匹配算法的实现方法，支持‘.’和‘*’两个特殊字符，其中‘.’可以匹配任意单个字符，而‘*’可以匹配其前面元素的零次或多次出现。文章通过递归方式实现了一个名为isMatch的功能，用于判断输入字符串是否完全符合给定的正则表达式。

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") → 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

分析与代码参考： http://leetcode.com/2011/09/regular-expression-matching.html

class Solution {
public:
	bool isMatch(const char *s, const char *p) {
		if (*p == '\0') return *s == '\0';
		// next char is not '*': must match current character
		if (*(p + 1) != '*') {
			return ((*p == *s) || (*p == '.' && *s != '\0')) && isMatch(s + 1, p + 1);
		}
		// next char is '*'
		while ((*p == *s) || (*p == '.' && *s != '\0')) {
			if (isMatch(s, p + 2)) return true;
			s++;
		}
		return isMatch(s, p + 2);
	}
};