Word Ladder

题目描述：
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 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.

分析：一道显而易见的BFS题目，使用BFS即可。

代码如下：

class Solution {public:struct data{    int x;    string y;};bool check(string a,string b){    int num=0;    if(a.size()==b.size()){        for(int i=0;i<b.size();i++){            if(a[i]!=b[i])            num++;        }    }    return num==1;}int ladderLength(string beginWord, string endWord, vector<string>& wordList) {        queue<data>temp;        int len=wordList.size();        vector<int>vis(len,0);        data t;        t.x=1;        t.y=beginWord;        temp.push(t);        while(!temp.empty()){            data p=temp.front();            temp.pop();            if(p.y==endWord){                return p.x;            }            for(int i=0;i<len;i++){                if(!vis[i]&&check(p.y,wordList[i])){                    vis[i]=1;                    data k;                    k.x=p.x+1;                    k.y=wordList[i];                    temp.push(k);                }            }        }        return 0;    }};