Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, such that:
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:
其实看到这个题就有广搜的想法,刚做完 一道经典搜索题 :http://blog.csdn.net/huruzun/article/details/39234511
下面是我AC java代码,没有经过太多优化,可以通过leetcode 。
public class Solution { public int ladderLength(String start, String end, Set<String> dict) { if(start==null || end == null || start.equals(end) || start.length()!=end.length()){ return 0; } Queue<String> queue = new LinkedList<String>(); queue.offer(start); Map<String, Integer> DictMap = new HashMap<String, Integer>(); DictMap.put(start, 1); // 利用广搜,start 字符串从头到尾每个字符进行更换。 while(!queue.isEmpty()){ String interString = queue.poll(); int TrasformTimes = DictMap.get(interString); for(int i=0;i<interString.length();i++){ for(char c='a';c<='z';c++){ if(interString.charAt(i)==c){ continue; } StringBuilder sb = new StringBuilder(interString); sb.setCharAt(i, c); if(sb.toString().equals(end)){ return TrasformTimes+1; } // 如果字典里有,并且先前没有出现 if(dict.contains(sb.toString())&& !DictMap.containsKey(sb.toString())){ queue.add(sb.toString()); DictMap.put(sb.toString(), TrasformTimes+1); } } } } return 0; } }
各位有更好思路请赐教,感谢。
原文地址:http://blog.csdn.net/huruzun/article/details/39253015