Longest Substring with At Most Two Distinct Characters

This problem can be solved in a similar approach as the Longest Substring Without Repeating Characters. Just for reference, I post the solution of this one here.

public class LongestSubstring {
    public int lengthOfLongestSubstring(String s) {
        if (s == null || s.length() == 0)
            return 0;
        HashSet hs = new HashSet();
        int leftBound = 0;
        int maxLength = 0;
        for (int i = 0; i < s.length(); i++)
        {
            if (!hs.contains(s.charAt(i)))
            {
                hs.add(s.charAt(i));
                maxLength = Math.max(maxLength, i - leftBound + 1);
                continue;
            }
            while (leftBound < i && s.charAt(leftBound) != s.charAt(i))
            {
                hs.remove(s.charAt(leftBound));
                leftBound++;
            }
            leftBound++;
        }
        return maxLength;
        
    }
}

Similarly, I use a moving window approach. A hash set is used to store the two characters of the substring. If the hash set doesn't contain the character, we add it, and we keep tracking the length of the substring that contains the characters in the hash set. The leftBound tracks the start point of the substring.
The tricky part comes when the third character comes up (char3). We need to move our window now! There are two possible cases:

1. s.charAt(leftBound) equals the character before the third character (let's say char2) and
2.  s.charAt(leftBound) doesn't equal to the character before the third character (char1).

The problem is, we only want to remove char1 and preserve char2, because char2 and char3 may form a longer substring.
For the second case, I use String.contains() method to check if the substring from charAt(leftBound) to char3 contains char1, if it does, we move the leftBound toward right.
For the first case, I just move the leftBound until I find the first char1. Then we can apply the solution for the second case again!

Based on my limited test case, it gives me the correct answer. However, since I haven't found enough convincible solutions online, I am open to any corrections.

import java.util.HashSet;


public class LongestSubstring {
 public int longestSubstring(String s)
 {
  if (s == null || s.length() == 0)
   return 0;
  HashSet hs = new HashSet();
  int leftBound = 0;
  int maxLength = 0;
  String sub = "";
  for (int i = 0; i < s.length(); i++)
  {
   if (!hs.contains(s.charAt(i)))
   {
    if (hs.size() < 2)
     hs.add(s.charAt(i));
    else
    {
     while (s.charAt(leftBound) == s.charAt(i - 1))
      leftBound++;
     CharSequence tmp = String.valueOf(s.charAt(leftBound));
     hs.remove(s.charAt(leftBound));
     while (s.substring(leftBound, i).contains(tmp))
      leftBound++;
     hs.add(s.charAt(i));
    }
   }
   maxLength = Math.max(maxLength, i - leftBound + 1);
   if (maxLength > sub.length())
    sub = s.substring(leftBound, i + 1);
  }
  System.out.println(sub);
  return maxLength;
 }

}

Update 2015 - 03 - 03:
I realize if all characters in the string are letters, then we can use a bit mask instead of a set, this may save some space.

public int longestSubstring3(String s){
  if (s == null || s.length() == 0)
         return 0;
  s = s.toLowerCase();
  int mask = 0;
  int size = 0;
  int max = 0;
  int leftBound = 0;
  for (int i = 0; i < s.length(); i++){
   if(size < 2){
    mask |= (1 << (s.charAt(i) - 'a'));
    size++;
   } else{
    if((mask & (1 << (s.charAt(i) - 'a'))) == 0){
     while(leftBound < i && s.charAt(leftBound) == s.charAt(i - 1))
      leftBound++;
     String tmp = String.valueOf(s.charAt(leftBound));
     mask &= (~(1 << (s.charAt(leftBound) - 'a')));
     while(leftBound < i && s.substring(leftBound, i).contains(tmp))
      leftBound++;
     mask |= (1 << (s.charAt(i) - 'a'));
    }
   }
   max = Math.max(max, i - leftBound + 1);
  }
  return max;
 }