[leetcode]127. Word Ladder

题目链接：https://leetcode.com/problems/word-ladder/description/

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 transformed word must exist in the word list. Note that beginWord is not a transformed word.

For example,

Given:
beginWord = "hit"
endWord = "cog"
wordList = ["hot","dot","dog","lot","log","cog"]

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.
• You may assume no duplicates in the word list.
• You may assume beginWord and endWord are non-empty and are not the same.

UPDATE (2017/1/20):
The wordList parameter had been changed to a list of strings (instead of a set of strings). Please reload the code definition to get the latest changes.

class Solution {public:    int ladderLength(string beginWord, string endWord, vector<string>& wordList) {        unordered_set<string> wordDict(wordList.begin(),wordList.end());        queue<string> toVisit;        addNextWords(beginWord,wordDict,toVisit);        int dist=2;        while(!toVisit.empty())        {            int num=toVisit.size();            for(int i=0;i<num;i++)            {                string word=toVisit.front();                toVisit.pop();                if(word==endWord)                    return dist;                addNextWords(word,wordDict,toVisit);            }            dist++;        }        return 0;    }private:    void addNextWords(string word,unordered_set<string>& wordDict,queue<string>& toVisit)    {        wordDict.erase(word);        for(int p=0;p<(int)word.length();p++)        {            char letter=word[p];            for(int k=0;k<26;k++)            {                word[p]='a'+k;                if(wordDict.find(word)!=wordDict.end())                {                    toVisit.push(word);                    wordDict.erase(word);                }            }            word[p]=letter;        }    }};