127. Word Ladder

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 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.

typedef struct{    string myString;    int step;}Node;bool canChange(string first, string second){    int length = first.length();    int diff = 0;    for(int i = 0; i < length; i ++){        if(first[i] - second[i] != 0){            diff ++;            if(diff > 1)                return false;        }    }//    if(diff == 1)        return true;}class Solution{    public:    int ladderLength(string beginWord, string endWord, vector<string>& wordList){        vector<Node> path;        Node begin = {beginWord, 1};        path.push_back(begin);        int start = 0, end = 0;        int amount = wordList.size();        bool flag[amount] = {false};        while(start <= end){                Node temp = path[start];                for(int i = 0; i < amount; i ++){                    if(!flag[i]){                        if(canChange(temp.myString, wordList[i])){                            Node node = {wordList[i], temp.step + 1};                            path.push_back(node);                            if(node.myString == endWord)                                return node.step;                            flag[i] = true;                            end ++;                        }                    }                }                start ++;            }        Node endNode = path[end];        if(endNode.myString == endWord)            return endNode.step;        else            return 0;    }};