138_leetcode_Word Ladder

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.

1:注意特殊情况；2:采用BFS遍历，将当前的单词的每一位设置为a~z的单词，对每一种情况进行处理；3:当存在end的时候直接返回; 4:对转变后的每一个单词，并且不存在与map中的单词进行处理。

    int ladderLength(string start, string end, unordered_set<string> &dict)    {        if(start.length() == 0 || end.length() == 0 || dict.size() == 0 || start.length() != end.length())        {            return 0;        }                deque<string> myDeque;        unordered_map<string, int> myMap;                myDeque.push_back(start);        myMap[start] = 1;                while(!myDeque.empty())        {            string curWord = myDeque.front();            int distLength = myMap[curWord];                        myDeque.pop_front();                        for(int i = 0; i < (int)curWord.length(); i++)            {                string tmpWord = curWord;                for(char j = 'a'; j <= 'z'; j++)                {                    if(tmpWord[i] == j)                    {                        continue;                    }                    else                    {                        tmpWord[i] = j;                        if(strcmp(tmpWord.c_str(), end.c_str()) == 0)                        {                            return distLength + 1;                        }                                                if(dict.find(tmpWord) != dict.end() && myMap.find(tmpWord) == myMap.end())                        {                            myDeque.push_back(tmpWord);                            myMap[tmpWord] = distLength + 1;                        }                    }                }            }        }        return 0;    }