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