Word Ladder 2015年6月3日
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.
Runtime:628ms
hit->hot->dot->dog->cog
hit->hot->lot->log->dog->cog
采用队列的优势就在只会在头部删除,尾部添加。
在队列中存在两元素 dot lot 时,dot启动查找匹配找到dict中存在的元素dog,并将其添加到队列中,因为队列的特性,dog会添加在lot之后
下一轮会先取出lot,这样就保证了每轮查找不同分支之间平行进行,最先return的就是最短的transformation
public class Solution { public int ladderLength(String beginWord, String endWord, Set<String> wordDict) { if (wordDict == null || wordDict.size() == 0) { return 0; } Queue<String> queue = new LinkedList<String>(); queue.offer(beginWord); wordDict.remove(beginWord); int length = 1; while(!queue.isEmpty()) { int count = queue.size(); for(int i=0; i<count; i++) { //Retrieves and removes the head of this queue, or returns null if this queue is empty 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(endWord)) { return length +1; } if(wordDict.contains(tmp)) { queue.offer(tmp); wordDict.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); } }
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