Word Ladder（LintCode）

Word Ladder

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

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.

用BFS做，字符串只相差一个字符的就当作有边。

 public class Solution {
     /**
       * @param start, a string
       * @param end, a string
       * @param dict, a set of string
       * @return an integer
       */
     public int ladderLength(String start, String end, Set<String> dict) {
         Queue q = new LinkedList();
         dict.add(end);
         int reslut = 0;

         //转成List方便标记,其实也可以将搜索过的String从Set中移除
         List<String> list = new ArrayList<String>(dict);
         int[] flag = new int[list.size()];
         q.add(start);
         int n = 1;
         int x = 0;
         while(q.size() > 0) {
             String y = (String)q.remove();
             n--;
             for (int i = 0;i<list.size() ;i++ ) {
                 if(end.equals(y)) return reslut+1;
                 if (judge(list.get(i),y) == 1 && flag[i] == 0) {
                     x++;
                     q.add(list.get(i));
                     flag[i] = 1;
                 }
             }
             if (n == 0) {
                 n = x;
                 x = 0;
                 reslut++;
             }
         }

         return 0;
     }

     public int judge(String a,String b) {
         int num = 0;
         char[] acs = a.toCharArray();
         char[] bcs = b.toCharArray();
         for (int i = 0;i<acs.length ;i++ ) {
             if(acs[i] != bcs[i]) num++;
         }
         return num;
     }
 }