Given two words (beginWord and endWord), and a dictionary's word list, find the length of shortest transformation sequence frombeginWord 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.

Note:

• Return 0 if there is no such transformation sequence.
• All words have the same length.
• All words contain only lowercase alphabetic characters.

每个单词每次只能变动一个字母变到字典中的其他单词，问从beginword变到endword最少需要多少步？

BFS，维护一个vis，表示已经到达过的单词，这样下次可以遇到可以不用搜索。

第一次提交超时了，以为BFS写搓了，调了半天发现是判断相邻单词的步骤超时。

原来版本的相邻词判断是这样的

bool judge(string a,string b)
{
bool flag=false;
for(int i=0;i<a.size();i++)
{
if(a[i]!=b[i])
{
if(flag) return false;
flag=true;
}
}
return flag;
}
for(unordered_set<string>::iterator it=wordList.begin();it!=wordList.end();it++)
{
unordered_set<string>::iterator it2=it;
for(it2++;it2!=wordList.end();it2++)
if(judge(*it,*it2))
{
mp[*it].push_back(*it2);
mp[*it2].push_back(*it);
}
}怎么样？遍历所有单词组合判断他们是否相邻，复杂度n*n*len(word),

改成下面的做法过了，

对于每一个词，枚举每一位的变化为其他字母的情况，判断这个情况是否在集合中，复杂度n*len(word)*26*1，确实要快上一些，学到了！

class Solution {
public:
int ladderLength(string beginWord, string endWord, unordered_set<string>& wordList) {
unordered_map<string,vector<string>> mp;
wordList.insert(beginWord);
wordList.insert(endWord);
for(unordered_set<string>::iterator it=wordList.begin();it!=wordList.end();it++)
{
string cc=*it;
for(int i=0;i<cc.size();i++)
{
string c=*it;
for(char j='a';j<='z';j++)
{
c[i]=j;
if(wordList.find(c)!=wordList.end())
{
mp[*it].push_back(c);
}
}
}
}
queue<string> q,q2;
unordered_map<string,bool> vis;
q.push(beginWord);
vis[beginWord]=true;
int res=0,flag=1;
while(!q.empty() || !q2.empty())
{
if(!q.empty())
{
string temp=q.front();
q.pop();
if(temp==endWord)
{
res=flag;
break;
}
for(int i=0;i<mp[temp].size();i++)
{
string tt=mp[temp][i];
if(vis[tt]) continue;
vis[tt]=true;
q2.push(tt);
}
}
else
{
flag++;
swap(q,q2);
}
}

return res;
}
};