word ladder

链接：http://leetcode.com/onlinejudge#question_126

原题：

Given two words (start and end), and a dictionary, find all shortest transformation sequence(s) 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"]

Return

  [    ["hit","hot","dot","dog","cog"],    ["hit","hot","lot","log","cog"]  ]

Note:

• All words have the same length.
• All words contain only lowercase alphabetic characters.

思路：

其实解题思路和Word Ladder完全一样，BFS，但是麻烦的是要返回所有的路径。

所以没办法，只能把每个单词所对应的前驱单词记录下来，当然有可能有多个，那么

就用一个vector<string>存储好，有这些记录就可以重构路径了。

代码：

[cpp] view plain copy

1. class Solution {  
2. public:  
3.     vector<vector<string> > findLadders(string start, string end, unordered_set<string> &dict) {  
4.         // Start typing your C/C++ solution below  
5.         // DO NOT write int main() function  
6.         pathes.clear();  
7.         dict.insert(start);  
8.         dict.insert(end);  
9.         vector<string> prev;  
10.         unordered_map<string, vector<string> > traces;  
11.         for (unordered_set<string>::const_iterator citr = dict.begin();   
12.                 citr != dict.end(); citr++) {  
13.             traces[*citr] = prev;  
14.         }  
15.           
16.         vector<unordered_set<string> > layers(2);  
17.         int cur = 0;  
18.         int pre = 1;  
19.         layers[cur].insert(start);  
20.         while (true) {  
21.             cur = !cur;  
22.             pre = !pre;  
23.             for (unordered_set<string>::const_iterator citr = layers[pre].begin();  
24.                     citr != layers[pre].end(); citr++)  
25.                 dict.erase(*citr);  
26.             layers[cur].clear();  
27.             for (unordered_set<string>::const_iterator citr = layers[pre].begin();  
28.                     citr != layers[pre].end(); citr++) {  
29.                 for (int n=0; n<(*citr).size(); n++) {    
30.                     string word = *citr;    
31.                     int stop = word[n] - 'a';    
32.                     for (int i=(stop+1)%26; i!=stop; i=(i+1)%26) {    
33.                         word[n] = 'a' + i;    
34.                         if (dict.find(word) != dict.end()) {    
35.                             traces[word].push_back(*citr);  
36.                             layers[cur].insert(word);   
37.                         }    
38.                     }  
39.                 }  
40.             }  
41.             if (layers[cur].size() == 0)  
42.                 return pathes;  
43.             if (layers[cur].count(end))  
44.                 break;  
45.         }  
46.         vector<string> path;  
47.         buildPath(traces, path, end);  
48.   
49.         return pathes;  
50.     }  
51.   
52.     private:  
53.         void buildPath(unordered_map<string, vector<string> > &traces,   
54.                 vector<string> &path, const string &word) {  
55.             if (traces[word].size() == 0) {  
56.                 path.push_back(word);  
57.                 vector<string> curPath = path;  
58.                 reverse(curPath.begin(), curPath.end());  
59.                 pathes.push_back(curPath);  
60.                 path.pop_back();  
61.                 return;  
62.             }  
63.   
64.             const vector<string> &prevs = traces[word];  
65.             path.push_back(word);  
66.             for (vector<string>::const_iterator citr = prevs.begin();  
67.                     citr != prevs.end(); citr++) {  
68.                 buildPath(traces, path, *citr);  
69.             }  
70.             path.pop_back();  
71.         }  
72.   
73.         vector<vector<string> > pathes;  
74. };