Word Ladder

Question:

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.

//********** Hints ************

LZ认为leetcode里面有带word题干的题目都好难，此题也是其中之一，但是思路比较简单，用BFS即可：

一个Queue作为Breadth,一个Hashset作为判断visited

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Solution:

public class Solution {   public int ladderLength(String start, String end, HashSet dict) {       Set<String> visited = new HashSet<String>();       Queue<String> breadth = new LinkedList<String>();              breadth.offer(start);       visited.add(start);              int step = 1;       int count = 1;              while(count > 0){           while(count > 0){               char[] cur = breadth.poll().toCharArray();                          for(int i = 0; i < cur.length; i++){                   char temp = cur[i];                   for(char c = 'a'; c <= 'z'; c++){                        if(c == temp)                            continue;                        cur[i] = c;                        String s = new String(cur);                        if(s.equals(end))                            return (step + 1);                        if(!visited.contains(s) && dict.contains(s)){                            breadth.offer(s);                            visited.add(s);                        }                   }                   cur[i] = temp;               }               count --;           }                      step ++;           count = breadth.size();       }              return 0;    }}