Leetcode 10 regular expression match -

Leetcode10_regularExpressionMatch.cs

using System;

class Solution
{
    public static bool IsMatch(string text, string pattern) // "", "a*b*"
    {
      if(text == null || pattern == null) // false 
      {
        return false; 
      }
      
      var tLength = text.Length; // 0 
      var pLength = pattern.Length; // 4
      
      var dp = new bool[tLength + 1, pLength + 1]; // 1, 5
      
      dp[0, 0] = true; // 
      
      for(int i = 1; i < tLength + 1; i++)
      {
        dp[i,0] = false; 
      }
      
      for(int i = 1; i < pLength + 1; i++)
      {
        dp[0, i] = false; 
        
        var visit = pattern[i - 1]; 
        var isStar = visit == '*'; 
        if(isStar)
        {
          if(i - 2 >= 0 )
          {
            dp[0, i] = dp[0, i - 2]; 
          }
        }
      }
      
      for(int row = 1; row < tLength + 1; row++)
      {
        for(int col = 1; col < pLength + 1; col++)
        {
          var visitChar   = text[row - 1]; 
          var patternChar = pattern[col - 1];
          
          var isStar = patternChar == '*'; 
          if(!isStar)
          {
            var isEqual = patternChar == '.' || visitChar == patternChar;             
             
            if(isEqual)
            {
              dp[row, col] = dp[row - 1, col - 1]; 
            }            
          }
          else 
          {                            //   zero time          one time           more then 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)
    {
    }
}

/*
base case: 

"", "a*b*c*", OK 
"abc", "",    false 
  
  dp[textLength + 1, patternLength + 1]
   ""  "a*"  "a*b*"   
   ------------------------
""  1 0  1  0  1
"a" 0 
  
            (i, j)
  
  if current char is not *, dp[i,j] =    
     dp[i - 1, j - 1] if text[i - 1] == pattern[i - 1] || pattern[i - 1] = '.'
     0                if text[i - 1] == pattern[i - 1]
     
  if current char is *, b* 
    match 0 time,        1 time                  or more than 1 time 
    dp[i, j - 2]  ||    dp[i, j - 1]            ||  dp[i - 1, j]    
*/