Word Ladder

最新推荐文章于 2024-09-08 21:02:59 发布

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本文探讨了如何从起始单词逐步变换至目标单词，确保每次仅改变一个字母且每步变化后的单词必须存在于字典中。通过实现一个算法，我们能够找到最短的转换路径长度，如果不存在这样的路径则返回0。

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Given two words (beginWord and endWord), and a dictionary, find the length of

shortest transformation sequence from beginWord to endWord, 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.

从前向后将字符串的字母替换，直至最后得到想要的字符串。

public int ladderLength(String start, String end, Set<String> dict) {
	if (dict == null || dict.size() == 0) 
		return 0;
	Queue<String> queue = new LinkedList<String>();
	queue.offer(start);
	dict.remove(start);
	int length = 1;
	while(!queue.isEmpty()) {
		int count = queue.size();
		for (int i = 0; i<count; i++){
			String current = queue.poll();
			for (char c = 'a'; c <= 'z'; c++) {
				for (int j=0; j < current.length(); j++) {
					if (c == current.charAt(j)) 
						continue;
					String tmp = replace(current, j, c);
					if (tmp.equals(end)) 
						return length + 1;
					if (dict.contains(tmp)){
						queue.offer(tmp);
						dict.remove(tmp);
					}
				}
			}
		}
		length++;
	}
	return 0;
}
private String replace(String s, int index, char c) {
	char[] chars = s.toCharArray();
	chars[index] = c;
	return new String(chars);
}