Leetcode 10 - regular expression match - dynamic programming

Leetcode10_regularExpressionMatch_Feb11_10AM.cs

using System;

class Solution
{
    public static bool IsMatch(string text, string pattern)
    {
      if(text == null || pattern ==  null)
      {
        return false; 
      }
      
      var tLength = text.Length; 
      var pLength = pattern.Length; 
      
      var dp = new bool[tLength + 1, pLength + 1]; 
      
      // set base case first 
      dp[0,0] = true; ; 
      
      // first row
      for(int col = 1; col < pLength + 1; col++)
      {
        var pChar = pattern[col - 1]; 
        var isStar = pChar == '*'; 
        if( col >= 2 && isStar)   // "" matches "a*" or "a*b*"
          dp[0, col] = dp[0, col - 2]; 
      }
      
      for(int row = 1; row < tLength + 1; row++)
      {
        for(int col = 1; col < pLength + 1; col++)
        {
          var tChar = text[row - 1]; 
          var pChar = pattern[col - 1]; 
          var isStar = pChar == '*';
          var isDot  = pChar == '.';
          
          if(!isStar)
          {
            dp[row, col] = (isDot || (tChar == pChar)) && dp[row - 1, col - 1]; // bug fixed
          }
          else 
          {
            // it is star
            //                 zero time,                     one time          or more than one time 
            dp[row, col] = (col >=2 || dp[row, col - 2]) || dp[row, col - 1] || dp[row - 1, col]; 
          }          
        }
      }
      
      return dp[tLength, pLength]; 
    }

    static void Main(string[] args)
    {

    }
}
/*
   b*  repeat zero time, one time and more than one time
   
   "",
   b
   bb
   bbbb
   
   acd with ab*c.
          0    1     2     3      4        5          pattern 
         ""   "a"  "ab"  "ab*"  "ab*c"  "ab*c."
         ----------------------------------------
 0  ""    T    F    F     F       F       F
 1  "a"   F    T    F     T(0)    F       F
 2  "ac"  F    
 3  "acd" F                              dp(3,5)=?
    text 
     
     base case:  "", dp[0,0] = true, dp[row,0] = false;
              text empty, "" match "a*" and "a*b*"
     
     inductive step: 
            discuss two case: 
            * involved - match zero, match one or more than one
            compare two key or if pattern key is wild char .
    */