127. Word Ladder(M)

127. Word LadderGiven two words (beginWord and endWord), and a dictionary‘s word list, find the length of shortest transformation sequence from beginWord to endWord, such that:

Only one letter can be changed at a time.
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.
UPDATE (2017/1/20):
The wordList parameter had been changed to a list of strings (instead of a set of strings). Please reload the code definition to get the latest changes.

• Difficulty:Medium
• Total Accepted:129.6K
• Total Submissions:667.5K
• Contributor:LeetCode

 1 class Solution {
 2 public:
 3     bool cmpwrd(string str1, string str2)
 4     {
 5         size_t strlen = str1.size();
 6         int cnt = 0;
 7         for (int i = 0; i < strlen; ++i)
 8         {
 9             if(str1[i] == str2[i]) ++cnt;
10         }
11         if(cnt == (strlen - 1)) return true;
12         return false;
13     }
14
15     int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
16         if(find(wordList.begin(), wordList.end(), endWord) == wordList.end()) return 0;
17         size_t wLlen = wordList.size();
18         vector<bool> visited(wLlen, false);
19         int sum = 1;
20         queue<string> qs;
21         qs.push(beginWord);
22         qs.push("#"); // level flag
23
24         while (!qs.empty())
25         {
26             string tmpstr = qs.front();
27             qs.pop();
28             if(tmpstr == "#")
29             {
30                 if(!qs.empty())
31                 {
32                     qs.push(tmpstr);
33                     tmpstr = qs.front();
34                     qs.pop();
35                     ++sum;
36                 }
37                 else return 0;
38             }
39
40             if(tmpstr == endWord) return sum;
41
42             //seek for all the possible next node
43             for (int j = 0; j < wLlen; ++j)
44             {
45                 if(!visited[j] && cmpwrd(tmpstr, wordList[j]))
46                 {
47                     if(tmpstr == endWord)
48                     {
49                         //cout << wordList[j] << "\n" << endWord << endl;
50                         return (++sum);
51                     }
52                     //cout << wordList[j] << " ";
53                     visited[j] = true;
54                     qs.push(wordList[j]);
55                 }
56             }
57             //cout << endl;
58         }
59         return sum;
60     }
61 };

12.21%  Runtime: 806 ms 39 / 39 test cases passed

 1 class Solution {  //by kaishen2
 2 public:
 3     int ladderLength(string beginWord, string endWord, vector<string>& wordDict) {
 4         unordered_set<string> word_dict_;
 5         for (auto &word : wordDict) word_dict_.insert(word);
 6         if (word_dict_.find(endWord) == word_dict_.end()) return 0;
 7         else word_dict_.erase(endWord);
 8         unordered_set<string> q1, q2, temp;
 9         q1.insert(beginWord);
10         q2.insert(endWord);
11         int count = 0;
12         while (!q1.empty() && !q2.empty()) {
13             ++count;
14             if (q1.size() > q2.size()) swap(q1, q2);
15             temp.clear();
16             for (auto word_ : q1) {
17                 for (int j = 0; j < word_.size(); ++j) {
18                     char hold = word_[j];
19                     for (char i = ‘a‘; i <= ‘z‘; ++i) {
20                         word_[j] = i;
21                         if (q2.find(word_) != q2.end())  return count + 1;
22                         if (word_dict_.find(word_) != word_dict_.end()) {
23                             word_dict_.erase(word_);
24                             temp.insert(word_);
25                         }
26                     }
27                     word_[j] = hold;
28                 }
29             }
30             swap(q1, temp);
31         }
32         return 0;
33     }
34 };

93.07% 35ms