Word Ladder leetcode

题目链接：http://oj.leetcode.com/problems/word-ladder/

Word Ladder

 Total Accepted: 9795 Total Submissions: 55989My Submissions

Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, such that:

1. Only one letter can be changed at a time
2. 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.

题目不难，就是用广搜，搜出最快到达的路径就成了，不过在搜的过程中，有两点要注意：

1.要有个变量来记录层数，以便于最后的输出。

2.由于‘a’到'z'就26个字母而已，而且每次只变一个，状态较少，而且那么多单词中只有少数符合要求，所以对原始单词进行修改，然后查看是否在dict中，即可。

code:

class Solution {public:    int ladderLength(string start, string end, unordered_set<string> &dict)     {        if(start.size()==0||(start.size()!=end.size()))            return 0;        if(start==end)            return 1;        queue<pair<string,int> > q;        q.push(make_pair(start,1));        dict.erase(start);        while(!q.empty())        {            string tmp=q.front().first;            int lev=q.front().second;            q.pop();            for(int i=0;i<tmp.size();i++)            {                char t=tmp[i];                for(int j=0;j<26;j++)                {                    if(t=='a'+j)                        continue;                    tmp[i]='a'+j;                    if(tmp==end)                    {                        return lev+1;                    }                    if(dict.count(tmp)>0)                    {                        q.push(make_pair(tmp,lev+1));                        dict.erase(tmp);                    }                }                tmp[i]=t;            }        }        return 0;        }};