Leetcode|Word Ladder

Given two words (beginWord and endWord), and a dictionary, 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 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.

此题采用广度优先搜索，因为每个单词的变换只是遍历每一位，然后更换其他25个字母。如果新的单词在字典里面而且之前没有用过（用map记录），就加入进来。
map的value表示给单词是第几步。key值表示单词。

int ladderLength(string beginWord, string endWord, unordered_set<string>& wordDict) {        queue<string> cur;        map<string,int> used;        used[beginWord]=1;        cur.push(beginWord);        while(!cur.empty())        {            string word=cur.front();            cur.pop();            if(word==endWord) break;            for(int i=0;i<word.size();i++){                string tmp=word;                for(char c='a';c<='z';c++){                    if(c==word[i]) continue;                    tmp[i]=c;                    if(wordDict.count(tmp)!=0&&used.find(tmp)==used.end()){                        used[tmp]=used[word]+1;                        cur.push(tmp);                    }                }            }        }        return used.find(endWord)==used.end()?0:used[endWord];    }