CODE 7: Word Ladder

最新推荐文章于 2019-01-11 00:33:45 发布

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最新推荐文章于 2019-01-11 00:33:45 发布

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分类专栏： leetcode

本文链接：https://blog.youkuaiyun.com/luzheqi/article/details/11584729

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本文探讨了如何从起始单词到目标单词的最短转换路径问题，仅通过一次字母改变，并确保每一步都存在于词汇表中。示例中，从hit到cog的最短路径为hit->hot->dot->dog->cog，路径长度为5。

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

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.equals(end)) {
			return 0;
		}
		
		Queue<String> que = new LinkedList<String>();
		que.offer(end);
		dict.add(start);
		int time = 1;
		int layer = que.size();
		while (!que.isEmpty()) {
			String now = que.poll();
			layer--;
			if (now.equals(start)) {
				return time;
			}
			isInDic(dict, now, que);

			if (layer == 0 && !que.isEmpty()) {
				layer = que.size();
				time++;
			}
		}
		return 0;
	}

	private void isInDic(HashSet<String> dict, String word, Queue<String> que) {
		for (int i = 0; i < word.length(); i++) {
			StringBuilder sb = new StringBuilder(word);
			for (char c = 'a'; c <= 'z'; c++) {
				sb.setCharAt(i, c);
				String t = sb.toString();
				if (dict.contains(t)) {
					dict.remove(t);
					que.offer(t);
				}
			}

		}
	}

Konw Something:

1. StringBuilder has a method named setCharAt(int arg0,char arg1).