Word Ladder

Problem:

Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, such that:

1. Only one letter can be changed at a time
2. 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.

我借用了dijkstra算法的思想。与dijkstra算法不同的是，每次对邻接点进行松弛后不用找最小值，因为步长只会是1，所有能够到达的节点都已是最短路径。如果字典的容量是n，单词的长度是m，算法的时间复杂度为O(26*n*m)。

Solution:

public class Solution {
HashSet<String> dic;
    public int ladderLength(String start, String end, Set<String> dict) {
        if(dict==null||start==null||end==null||dict.size()==0||start.equals("")||end.equals(""))
         return 0;
        if(start.equals(end))
             return 1;
        dic = new HashSet<>(dict);
        dic.add(end);
        int pathLen = 1;
        LinkedList<String> curStrings = new LinkedList<>();
        curStrings.add(start);
               
        while(!curStrings.isEmpty())
        {
             int count = curStrings.size();
             pathLen++;
             while(count-->0)
             {
                 String str = curStrings.poll();
                 char[] chars = str.toCharArray();
                 for(int i=0;i<chars.length;i++)
                 {
                     char temp = chars[i];
                     chars[i] = 'a'-1;
                     while(chars[i]++<'z')
                     {
                         if(chars[i]==temp)
                             continue;
                         String s = new String(chars);
                         if(s.equals(end))
                             return pathLen;
                         if(dic.contains(s))
                         {
                             dic.remove(s);
                             curStrings.offer(s);
                         }
                     }
                     chars[i] = temp;
                 }
             }
        }
        return 0;
    }
}