题目描述：
Given two words (beginWord and endWord), and a dictionary's word list, find the length of shortest transformation sequence from beginWord to endWord, such that:
Only one letter can be changed at a time
Each intermediate word must exist in the word list
For example,
Given:
beginWord = "hit"
endWord = "cog"
wordList = ["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.
思路：
就是给定1个单词数组(wordList)和起始单词(beginWord)以及终止单词(endWord)，求从beginWord到endWord的最小变化过程：
对于beginWord[0...n]，每次只能变一个字符得到新单词newWord，并且newWord必须在wordList中存在才能进行下一次变化。
本题可以理解为，每个单词代表一个节点，每次求出“梯子节点”（改变1个字符后在WordList中存在），对所有的梯子节点继续遍历，步骤数+1。求从beginWord到endWord的最小变化步骤数。
由于目标是步数最小化，因此本题是一个BFS问题。
1.由于题目已经说明，所有单词为'a'-'z'字符组成，因此每次单词的变化可以为：将word[0...n)的每个字符分别替换为不等于word[i]的字符'a'-'z'，得到新单词newWord，如果newWord在词典中存在，添加到‘梯子节点’集合中。
2.从beginWord开始，不断的查找下一个梯子节点，循环判断梯子节点是否为endWord，如果不是则准备下一轮的梯子节点集合。
3.BFS一共跑的循环次数就等于最小步骤数。
实现代码：public class Solution {
public int LadderLength(string beginWord, string endWord, ISet<string> wordList)
{
if(beginWord.Length == 1 )
{
return beginWord[0] != endWord[0] ? 2 : 0;
}
wordList.Remove(beginWord);
wordList.Add(endWord);
var matchedWords = new HashSet<string>();
matchedWords.Add(beginWord);
var found = false;
var result = new List<ISet<string>>(){matchedWords};
while(matchedWords.Count > 0 && !found){
var nextRound = new HashSet<string>();
foreach(var w in matchedWords){
if(w == endWord){
found = true;
break;
}
else{
var r = NextLadderWords(w, ref wordList);
foreach(var w1 in r){
nextRound.Add(w1);
}
}
}
matchedWords = nextRound;
if(!found){
result.Add(nextRound);
}
}
return found ? result.Count : 0;
}
private ISet<string> NextLadderWords(string word,ref ISet<string> words){
var ret = new HashSet<string>();
var len = word.Length;
for(var i = 0;i < len; i++){
var c = (int)word[i];
for(var j = 'a'; j <= 'z'; j++){
if(c != j){
var newWord = word.ToCharArray();
newWord[i] = (char)j;
var s = new string(newWord);
if(words.Contains(s)){
ret.Add(s);
words.Remove(s);
}
}
}
}
return ret;
}
}