Finding a Palindrome

6 Comments

• Nice refactoring! :-)
  
  
  
  As a TDDer, expressing the rules with Asserts just makes so much sense - nice to see others think the same!
  
  
  
  I have always thought of palindromes as being more loose, such as:
  
  
  
  Madam I'm Adam
  
  
  
  Although I guess this would simply require same-casing everything and removing whitespace and punctuation ...
  

• Uhm, how about this?
  
  
  
  private bool IsPalindrome(string text)
  
  {
  
  if(text==null) return false;
  
  text=text.ToLower();
  
  string backwards=Microsoft.VisualBasic.StrReverse(text);
  
  return (backwards==text);
  
  }
  

• 
  "X" == "X" ? Or "X".Equals( "X" ) ?
  

• As per the original algorithm
  
  just modifying a little, i like the idea of going half way than using string operations which are expensive
  
  
  
  char[] a = text.ToCharArray();
  
  bool b = true;
  
  int l = a.length;
  
  int i =0;
  
  while ( b && i < l/2 )
  
  {
  
  if ( !(a[i] == a[l-i-1]) ) b=false;
  
  i++;
  
  }
  
  return b;
  
  
  
  

• You should probably get rid of white space first. Generally people don't consider white space in palindromes.
  

• Mani's solution is good, but why create a character array unnecessarily, and why keep looping when you've established they're different? The string class already has an indexer so you can do:
  
  if (text[i] == test[l-i-1])
  
  
  
  You could generalize it to ignore whitespace and punctuation (air code) ...
  
  
  
  int firstIndex = 0;
  
  int lastIndex = text.Length-1;
  
  while(firstIndex < lastIndex)
  
  {
  
  // Ignore punctuation and whitespace
  
  if (Char.IsPunctuation(text[firstIndex]) || Char.IsWhiteSpace(text[firstIndex]))
  
  {
  
  firstIndex++;
  
  continue;
  
  }
  
  if (Char.IsPunctuation(text[lastIndex]) || Char.IsWhiteSpace(text[lastIndex]))
  
  {
  
  lastIndex--;
  
  continue;
  
  }
  
  if (Char.ToLower(text[firstIndex]) != Char.ToLower(text[lastIndex])) return false;
  
  firstIndex++;
  
  lastIndex--;
  
  }
  
  return true;