leetcode Word Ladder

废话少说，先上题：

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.

 这题本人的思路是利用bfs（广度优先搜索）来做，比如第一次单词是fit，先找出与fit距离为一的单词，若等于end，则返回，若不是则将该单词加入队列中，重复上述的动作，因为每次扫描都是距离相等的点的集合，若找到则返回，找不到则返回-1，但是，尼玛，爷做的怎么就不能AC呢，fuck，上不能AC的代码：

class Solution {public:queue<string> que;map<string, bool> mapping;bool add(string begin, string end, unordered_set<string>& dict){for (int i = 0; i < begin.size(); i++){for (char j = 'a'; j <= 'z'; j++){string tmp = begin;if (tmp[i] != j){tmp[i] = j;if (!mapping[tmp]){if (dict.find(tmp) != dict.end()){que.push(tmp);if (tmp == end){return true;}mapping[tmp] = true;}}}}}return false;}int bfs(string start, string end, unordered_set<string>& dict){if (add(start, end, dict)){return 2;}int count = 3;int tail = que.size();if (tail == 0){return 0;}while (true){int begin = 0;while (begin < tail){string s = que.front();que.pop();if (add(s, end, dict)){return count;}begin++;}count++;tail = que.size();if (tail == 0){return 0;}}}int ladderLength(string start, string end, unordered_set<string> &dict){for (auto i = dict.begin(); i != dict.end(); i++){mapping[*i] = false;}if (start == end){return 0;}return bfs(start, end, dict);}};