10. 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

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This problem has a typical solution using Dynamic Programming. We define the state P[i][j] to be true if s[0..i) matches p[0..j)and false otherwise. Then the state equations are:

1. P[i][j] = P[i - 1][j - 1], if p[j - 1] != '*' && (s[i - 1] == p[j - 1] || p[j - 1] == '.')
   ;
2. P[i][j] = P[i][j - 2], if p[j - 1] == '*' and the pattern repeats for 0 times;
3. P[i][j] = P[i - 1][j] && (s[i - 1] == p[j - 2] || p[j - 2] == '.'), if p[j - 1] == '*' and the pattern repeats for at least 1 times.

class Solution {
public:
	bool isMatch(string s, string p) {
		int m = s.size(), n = p.size();
		vector<vector<bool>> dp(m + 1, vector<bool>(n + 1,false));
		dp[0][0] = true;

		for (int i = 0; i <= m; i++){
			for (int j = 1; j <= n; j++){
				if (p[j - 1] == '*'){
					dp[i][j] = dp[i][j - 2] ||
						(i > 0 && dp[i - 1][j] && (s[i - 1] == p[j - 2] || p[j - 2] == '.'));
				}
				else{
					if (i > 0 && (s[i-1] == p[j-1] || p[j-1] == '.')){
						dp[i][j] = dp[i - 1][j - 1];
					}
				}
			}
		}
		return dp[m][n];
	}
};
 

44. Wildcard Matching

Implement wildcard pattern matching with support for '?' and '*'.

'?' Matches any single character.
'*' Matches any sequence of characters (including the empty sequence).

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", "*") → true
isMatch("aa", "a*") → true
isMatch("ab", "?*") → true
isMatch("aab", "c*a*b") → false

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這個跟上面的區別是，這裡*是萬用字元可以匹配任何字元，而不是它前面的那個字元。

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