Word Ladder

题目：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：

定义一个队列，以及一个hash表，hash里面存放当前字符和step数。

最一开始，先存入start，数一下，从当前start，变换每一个字符，这里面一共改变26*size次数：

如果改变之后的字符存在并且不曾在visited中出现过，插入当前字符串，step加1，

如果改变的字符串等于end，直接返回。

代码：

class Solution {
public:
int ladderLength(string start, string end, unordered_set<string> &dict) {
queue<pair<string,int> >q;
unordered_set<string> visited;

q.push(make_pair(start,1));
visited.insert(start);

//开始判断

while(!q.empty()){
string curStr=q.front().first;
int curStep=q.front().second;
q.pop();

for(int i=0;i<curStr.size();i++){
string tmp=curStr;
for(int j=0;j<26;j++){
tmp[i] = j+'a';
if(tmp==end)    return curStep+1;
if(visited.find(tmp)==visited.end()&&dict.find(tmp)!=dict.end()){
//为了避免"hot" "dog" ["hot","dog"] 这种情况下，程序不动，一直在运行
visited.insert(tmp);
q.push(make_pair(tmp,curStep+1));
}
}
}
}
return 0;
}
};