【LeetCode】44. Regular Expression Matching

题目描述(Hard)

Given an input string (s) and a pattern (p), 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).

Note:

• s could be empty and contains only lowercase letters a-z.
• p could be empty and contains only lowercase letters a-z, and characters like . or *.

题目链接

https://leetcode.com/problems/longest-palindromic-substring/description/

Example 1:

Input:
s = "aa"
p = "a"
Output: false
Explanation: "a" does not match the entire string "aa".

Example 2:

Input:
s = "aa"
p = "a*"
Output: true
Explanation: '*' means zero or more of the precedeng element, 'a'. Therefore, by repeating 'a' once, it becomes "aa".

Example 3:

Input:
s = "ab"
p = ".*"
Output: true
Explanation: ".*" means "zero or more (*) of any character (.)".

Example 4:

Input:
s = "aab"
p = "c*a*b"
Output: true
Explanation: c can be repeated 0 times, a can be repeated 1 time. Therefore it matches "aab".

Example 5:

Input:
s = "mississippi"
p = "mis*is*p*."
Output: false

算法分析

【剑指】19.正则表达式匹配

提交代码：

class Solution {
public:
	bool isMatch(string s, string p) {
		return isMatchCore(s.c_str(), p.c_str());
	}

	bool isMatchCore(const char *str, const char *pattern)
	{
		if (*str == '\0' && *pattern == '\0') return true;
		if (*str != '\0' && *pattern == '\0') return false;

		if (*(pattern + 1) == '*')
		{
			if (*str == *pattern || (*pattern == '.' && *str != '\0'))
				return isMatchCore(str + 1, pattern)
				|| isMatchCore(str, pattern + 2)
				|| isMatchCore(str + 1, pattern + 2);
			else
				return isMatchCore(str, pattern + 2);
		}

		if (*str == *pattern || (*pattern == '.' && *str != '\0'))
			return isMatchCore(str + 1, pattern + 1);

		return false;
	}
};

测试代码：

// ====================测试代码====================
void Test(const char* testName, string str, string p, bool expected)
{
	if (testName != nullptr)
		printf("%s begins: \n", testName);

	Solution s;
	bool result = s.isMatch(str, p);

	if(result == expected)
		printf("passed\n");
	else
		printf("failed\n");
}

int main(int argc, char* argv[])
{
	
	Test("Test1", string("aa"), string("a"), false);
	Test("Test2", string("aa"), string("a*"), true);
	Test("Test3", string("ab"), string(".*"), true);
	Test("Test4", string("aab"), string("c*a*b"), true);
	Test("Test5", string("mississippi"), string("mis*is*p*."), false);

	return 0;
}