Word Ladder

 Given two words (start and end), and a dictionary, find the length of shortest transformation sequence fromstart toend, such that:

1. Only one letter can be changed at a time
2. Each intermediate word must exist in the dictionary

For example,

Given:
start = "hit"
end = "cog"
dict = ["hot","dot","dog","lot","log"]

As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog",
return its length 5.

Note:

• Return 0 if there is no such transformation sequence.
• All words have the same length.
• All words contain only lowercase alphabetic characters.

	public int ladderLength(String start, String end, HashSet<String> dict) {
		// Start typing your Java solution below
		// DO NOT write main() function
		if (start.length() != end.length())
			return 0;
		if (start.equals(end))
			return 2;
		int path = 0;

		Queue<String> queue = new LinkedList<String>();
		queue.offer(start);
		HashSet<String> hitted = new HashSet<String>();
		hitted.add(start);
		int queue_counter = 1;
		int set_counter = 0;

		while (!queue.isEmpty()) {
			String s = queue.poll();
			queue_counter--;
			HashSet<String> valid_neighbor = nextHit(s, dict, hitted);
			set_counter = valid_neighbor.size();
			for (String str : valid_neighbor) {
				if (str.equals(end))
					return path + 2;
				else
					queue.offer(str);
			}
			if (queue_counter == 0) {
				path++;
				queue_counter = queue.size();
				// queue_counter = set_counter;
				// fixed a bug here.
				// set_counter is not suitable to be a queue_counter
			}
		}

		return 0;
	}

	private HashSet<String> nextHit(String s, HashSet<String> dict,
			HashSet<String> hitted) {
		HashSet<String> valid_neighbor = new HashSet<String>();
		for (int len = 0; len < s.length(); len++) {
			StringBuilder sb = new StringBuilder(s);
			for (char ch = 'a'; ch <= 'z'; ch++) {
				sb.setCharAt(len, ch);
				String test_case = sb.toString();
				if (dict.contains(test_case) && !hitted.contains(test_case)) {
					hitted.add(test_case);
					valid_neighbor.add(test_case);
				}
			}
		}
		return valid_neighbor;
	}