127. Word Ladder

1. 问题描述
   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.

   Note:
   Return 0 if there is no such transformation sequence.
   All words have the same length.
   All words contain only lowercase alphabetic characters.
   You may assume no duplicates in the word list.
   You may assume beginWord and endWord are non-empty and are not the same.

2. 解决思路
   找到当前字符串与它相差一个字符的字符串作为它的子节点，然后进行广搜即可。

3. 代码

class Solution {public:    int ladderLength(string beginWord, string endWord, vector<string>& wordList) {        vector<bool> isVisit(wordList.size(),0);        queue<string> que,next;        que.push(beginWord);        int count = wordList.size(), len = 0;        for (int i = 0; i < wordList.size(); ++i) {            if(beginWord == wordList[i]) {                isVisit[i] = true;                --count;                break;            }        }        while(!que.empty() || count) {            string tmp = que.front();            que.pop();            if (tmp == endWord)                return len+1;            for (int i = 0; i < wordList.size(); ++i) {                int new_count = 0;                for (int j = 0; j < beginWord.length(); ++j) {                    if(tmp[j] != wordList[i][j])                        ++new_count;                }                if (new_count == 1 && !isVisit[i]) {                    isVisit[i] = true;                    next.push(wordList[i]);                    --count;                }            }            //cout<<"next's size" << next.size() <<endl;            if (que.empty()) {                if (next.empty())                    return 0;                que = next;                next = queue<string>();                ++len;            }        }        //cout <<"count: "<<count<<endl;        return 0;    }};