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:
- P[i][j] = P[i - 1][j - 1], if p[j - 1] != '*' && (s[i - 1] == p[j - 1] || p[j - 1] == '.');
- P[i][j] = P[i][j - 2], if p[j - 1] == '*' and the pattern repeats for 0 times;
- 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.
Putting these together, we will have the following code.
1 class Solution { 2 public: 3 bool isMatch(string s, string p) { 4 int m = s.length(), n = p.length(); 5 vector<vector<bool> > dp(m + 1, vector<bool> (n + 1, false)); 6 dp[0][0] = true; 7 for (int i = 0; i <= m; i++) 8 for (int j = 1; j <= n; j++) 9 if (p[j - 1] == '*') 10 dp[i][j] = dp[i][j - 2] || (i > 0 && (s[i - 1] == p[j - 2] || p[j - 2] == '.') && dp[i - 1][j]); 11 else dp[i][j] = i > 0 && dp[i - 1][j - 1] && (s[i - 1] == p[j - 1] || p[j - 1] == '.'); 12 return dp[m][n]; 13 } 14 };