题目：

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.

题解：

这道题是套用BFS同时也利用BFS能寻找最短路径的特性来解决问题。

把每个单词作为一个node进行BFS。当取得当前字符串时，对他的每一位字符进行从a~z的替换，如果在字典里面，就入队，并将下层count++，并且为了避免环路，需把在字典里检测到的单词从字典里删除。这样对于当前字符串的每一位字符安装a~z替换后，在queue中的单词就作为下一层需要遍历的单词了。

正因为BFS能够把一层所有可能性都遍历了，所以就保证了一旦找到一个单词equals（end），那么return的路径肯定是最短的。

像给的例子 start = hit，end = cog，dict = [hot, dot, dog, lot, log]

按照上述解题思路的走法就是：

level = 1    hit   dict = [hot, dot, dog, lot, log]

ait bit cit ... xit yit zit ，  hat hbt hct ... hot ... hxt hyt hzt ，  hia hib hic ... hix hiy hiz

level = 2    hot  dict = [dot, dog, lot, log]

aot bot cot dot ...  lot ... xot yot zot，hat hbt hct ... hxt hyt hzt，hoa hob hoc ... hox hoy hoz

level = 3             wordQueue.add(start);
         
                      String word = wordQueue.poll();
             curnum--;
             
                                                                    wordunit[i] = j;
                     String temp =                      
                                                                                            wordQueue.add(temp);
                         nextnum++;
                         dict.remove(temp);
                     }
                 }
             }
             
                              curnum = nextnum;
                 nextnum = 0;
                 level++;
             }
         }
         
              }