Leetcode163: Word Ladder

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:

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

网上对该问题的解决无一例外都是按照下面的思路：将当前单词每一个字符替换成a~z的任意一个字符，然后判断是否在词典中出现。此时的复杂度是O(26*word_length)，当单词比较短时，这种方法的优势就体现出来了。

class Solution {public:    int ladderLength(string beginWord, string endWord, unordered_set<string>& wordList) {        if(beginWord.size() != endWord.size()) return 0;          if(beginWord.empty() || endWord.empty())return 0;                    queue<string> path;          path.push(beginWord);          int level = 1;          int count = 1;          wordList.erase(beginWord);          while(wordList.size() > 0 && !path.empty())          {              string curword = path.front();              path.pop();count--;              for(int i = 0; i < curword.size(); i++)              {                  string tmp = curword;                  for(char j='a'; j<='z'; j++)                  {                      if(tmp[i]==j)continue;                      tmp[i] = j;                      if(tmp==endWord)return level+1;                      if(wordList.find(tmp) != wordList.end()) path.push(tmp);                      wordList.erase(tmp);                  }              }              if(count==0)              {                  count = path.size();                  level++;              }          }          return 0;      }};