[LeetCode] Regular Expression Matching

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.

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 };

 

posted @ 2015-07-05 23:40  jianchao-li  阅读(408)  评论(0)  编辑  收藏  举报