[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：



每次只能改变一个字母。

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];
}

}
};