[LeetCode]10. Regular Expression Matching（自学留存）

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

Personal tips:这道题真的挺难的，使用递归算法，时间太长了，应该用dp改进，以下是递归的代码实现：

class Solution {
public:
	bool isMatch(string s, string p) {
		int i = 0, j = 0;
		if (p.empty()) return s.empty();
		if (p[j + 1] != '*')
		{
			if (s[i] == p[j]|| (p[j] == '.'&&s[i] != '\0')) return isMatch(s.substr(i + 1), p.substr(j + 1));
			else return false;
		}
		while (s[i]==p[j]||(p[j]=='.'&&s[i]!='\0'))
		{
			if (isMatch(s.substr(i), p.substr(j + 2)))  return true; 
			++i; 
		}
		return isMatch(s.substr(i), p.substr(j + 2));
	}
};

以下为参考LeetCode后的动态规划dp算法实现代码：

class Solution {
public:
	bool isMatch(string s, string p) {	
		if (s.empty()&&p.empty()) return true;
		if ( p.empty()) return false;
		vector<vector<bool>> dp(s.size() + 1, vector<bool>(p.size() + 1, false));
		dp[0][0] = true;
		for (int i = 0; i < s.size() + 1; ++i)
		{
			
			for (int j = 1; j < p.size() + 1; ++j)
			{
				if (i>0&&(s[i-1] == p[j-1] || p[j-1] == '.')) dp[i][j] = (dp[i - 1][j - 1]?true:dp[i][j]);
				if (p[j-1] == '*')
				{
					if (i>0&&(p[j - 1-1] == s[i-1]|| p[j - 1 - 1] == '.'))
					{
						dp[i][j] = dp[i][j - 1] || dp[i - 1][j]||dp[i][j-2];
					}
					else
					{
						dp[i][j] = dp[i][j - 2];
					}
				}
			}
		}
		return dp[s.size()][p.size()];		
	}
};