dp solution for regular matching

Author: applewjg

RegularExpressionMatching.h

@@ -1,7 +1,8 @@
 /*
  Author:     Annie Kim, [email protected]
+ Co-author:  King, [email protected]
  Date:       May 26, 2013
- Update:     Aug 19, 2013
+ Update:     Oct 26, 2014
  Problem:    Regular Expression Matching
  Difficulty: Hard
  Source:     http://leetcode.com/onlinejudge#question_10
@@ -22,6 +23,7 @@
  isMatch("aab", "c*a*b") ? true
 
  Solution: Both of the two solutions are from http://leetcode.com/2011/09/regular-expression-matching.html .
+ Solution 3: DP.
 */
 
 class Solution {
@@ -53,4 +55,23 @@ class Solution {
         else
             return *(p+1) == '*' && isMatch(s, p+2);
     }
+
+    bool isMatch_3(const char *s, const char *p) {
+         int l1 = strlen(s), l2 = strlen(p), k;
+         char ch1, ch2;
+         vector<vector<bool> > f(l1 + 1, vector<bool>(l2 + 1,false));
+         f[0][0] = true;
+         for (int i = 2; i <= l2; i ++)
+            if (*(p + i - 1) == '*') f[0][i] = f[0][i - 2];
+         for (int i = 1; i <= l1; i ++)
+            for (int j = 1; j <= l2; j ++) {
+                ch1 = *(s + i - 1); ch2 = *(p + j - 1);
+                if (ch2 != '*') f[i][j] = f[i - 1][j - 1] && (ch1 == ch2 || ch2 == '.');
+                else {
+                    f[i][j] = f[i][j - 2];
+                    if (*(s + i - 1) == *(p + j - 2) || *(p + j - 2) == '.') f[i][j] = f[i][j] | f[i - 1][j];
+                }
+            }
+         return f[l1][l2];
+    }
 };