Java for LeetCode 126 Word Ladder II 【HARD】

Given two words (start and end), and a dictionary, find all shortest transformation sequence(s) from start to end, such that:

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

For example,

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

Return

  [
    ["hit","hot","dot","dog","cog"],
    ["hit","hot","lot","log","cog"]
  ]
解题思路：

本题应该是目前遇到的最Hard的一道了，思路先按照Java for LeetCode 127 Word Ladder的代码进行dfs找到 ladderLength 然后以 ladderLength 为步长进行DFS，这里进行DFS需要从后往前（因为disMap是从前往后建立的，从前往后的话前期肯定有无数匹配，从后往前的话，只要匹配就是要找到alist）

JAVA实现如下：

static public List<List<String>> findLadders(String start, String end,Set<String> dict) {
		List<List<String>> result = new LinkedList<List<String>>();
		LinkedList<String> queue = new LinkedList<String>();
		HashMap<String, Integer> disMap = new HashMap<String, Integer>();
		queue.add(start);
		disMap.put(start, 1);
		int depth = -1;
		findDepth: while (!queue.isEmpty()) {
			String word = queue.poll();
			for (int i = 0; i < word.length(); i++) {
				StringBuilder sb = new StringBuilder(word);
				for (char ch = 'a'; ch <= 'z'; ch++) {
					sb.setCharAt(i, ch);
					String nextWord = sb.toString();
					if (nextWord.equals(end)) {
						depth = disMap.get(word) + 1;
						break findDepth;
					}
					if (dict.contains(nextWord)
							&& !disMap.containsKey(nextWord)) {
						queue.add(nextWord);
						disMap.put(nextWord, disMap.get(word) + 1);
					}
				}
			}
		}
		if (depth > 0)
			dfs(result, start, end, disMap, depth,new LinkedList<String>());
		return result;
	}

	static void dfs(List<List<String>> result, String start, String end,HashMap<String, Integer> disMap, int depth,List<String> alist) {
		alist.add(0, end);
		if (end.equals(start))
			result.add(new LinkedList<String>(alist));
		if (depth <= 1)
			return;
		String word = alist.get(0);
		for (int i = 0; i < word.length(); i++) {
			StringBuilder sb = new StringBuilder(word);
			for (char ch = 'a'; ch <= 'z'; ch++) {
				sb.setCharAt(i, ch);
				String nextWord = sb.toString();
				if (disMap.containsKey(nextWord)&&disMap.get(nextWord)==depth-1) {
					dfs(result, start, nextWord, disMap, depth - 1,alist);
					alist.remove(0);
				}
			}
		}
	}