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:

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.

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;}};