[LeetCode 127] Word Ladder

https://leetcode.com/problems/word-ladder/

http://www.lintcode.com/zh-cn/problem/word-ladder/

给出两个单词（start和end）和一个字典，找到从start到end的最短转换序列

比如：

1. 每次只能改变一个字母。
2. 变换过程中的中间单词必须在字典中出现。

样例

给出数据如下：

start = "hit"

end = "cog"

dict = ["hot","dot","dog","lot","log"]

一个最短的变换序列是 "hit" -> "hot" -> "dot" -> "dog" -> "cog"，

返回它的长度 5

注意

• 如果没有转换序列则返回0。
• 所有单词具有相同的长度。
• 所有单词都只包含小写字母。

// --------------------------- 

//  BFS non-recursive method

// ---------------------------

//

//    Using BFS instead of DFS is becasue the solution need the shortest transformation path.

//  

//    So, we can change every char in the word one by one, until find all possible transformation.

//

//    Keep this iteration, we will find the shorest path.

//

// For example:

//   

//     start = "hit"

//     end = "cog"

//     dict = ["hot","dot","dog","lot","log","dit","hig", "dig"]

//

//                      +-----+                  

//        +-------------+ hit +--------------+   

//        |             +--+--+              |   

//        |                |                 |   

//     +--v--+          +--v--+           +--v--+

//     | dit |    +-----+ hot +---+       | hig |

//     +--+--+    |     +-----+   |       +--+--+

//        |       |               |          |   

//        |    +--v--+         +--v--+    +--v--+

//        +----> dot |         | lot |    | dig |

//             +--+--+         +--+--+    +--+--+

//                |               |          |   

//             +--v--+         +--v--+       |   

//        +----> dog |         | log |       |   

//        |    +--+--+         +--+--+       |   

//        |       |               |          |   

//        |       |    +--v--+    |          |   

//        |       +--->| cog |<-- +          |   

//        |            +-----+               |   

//        |                                  |   

//        |                                  |   

//        +----------------------------------+   

//     

//     1) queue <==  "hit"

//     2) queue <==  "dit", "hot", "hig"

//     3) queue <==  "dot", "lot", "dig"

//     4) queue <==  "dog", "log" 

// 



class Solution {

public:

    int ladderLength(string beginWord, string endWord, unordered_set<string>& wordDict) {

        unordered_map<string, int> dis;

        dis.insert(make_pair(beginWord, 1));

        

        queue<string> q;

        q.push(beginWord);

        

        while (!q.empty()) {

            string word = q.front();

            q.pop();

            

            if (word == endWord) {

                break;

            }

            

            for (int i = 0; i < word.size(); i++) {

                string tmp = word;

                for (char c = 'a'; c <= 'z'; c++) {

                    tmp[i] = c;

                    if (wordDict.count(tmp) == 1 && dis.count(tmp) == 0) {

                        q.push(tmp);

                        dis.insert(make_pair(tmp, dis[word] + 1));

                    }

                }

            }

        }

        

        if (dis.count(endWord) == 0) {

            return 0;

        } else {

            return dis[endWord];

        }

        

    }

};