Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, such that:
- Only one letter can be changed at a time
- 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
思路：
这道题的思路应该是图和BFS。
最开始写错了，写成了图和DFS，导致总是超时。后来修改的时候增加了记录中间结果以及剪枝，但还总是超时。代码如下：
1 map<string, vector<string> > table; 2 map<string, bool> visit; 3 map<string, int> result; 4 int minAll; 5 void initializeTable(string s, unordered_set<string> &dict){ 6 int l = s.length(); 7 int i,n; 8 for(unordered_set<string>::iterator it = dict.begin(); it != dict.end(); it++){ 9 n = 0; 10 string t = *it; 11 for(i = 0; i < l; i++){ 12 if(s[i] != t[i]) 13 n++; 14 if(n>1) 15 break; 16 } 17 if(n == 1) 18 table[s].push_back(t); 19 } 20 } 21 int minLength(string start, string end, int step){ 22 if(start == end) 23 return 1; 24 if(table.find(start) == table.end()) 25 return 99999; 26 if(result.find(start) != result.end()) 27 return result[start]; 28 if(step >= minAll) 29 return 99999; 30 vector<string> next = table[start]; 31 visit[start] = true; 32 int curMin = 99999; 33 for(int i = 0; i < next.size(); i++){ 34 if(visit[next[i]]) 35 continue; 36 int t = 1 + minLength(next[i], end, step+1); 37 if(t < curMin) 38 curMin = t; 39 } 40 visit[start] = false; 41 result[start] = curMin; 42 if(minAll > step+curMin) 43 minAll = step+curMin; 44 return curMin; 45 } 46 int ladderLength(string start, string end, unordered_set<string> &dict) { 47 // IMPORTANT: Please reset any member data you declared, as 48 // the same Solution instance will be reused for each test case. 49 table.clear(); 50 initializeTable(start, dict); 51 initializeTable(end, dict); 52 dict.insert(end); 53 minAll = INT_MAX; 54 for(unordered_set<string>::iterator it = dict.begin(); it != dict.end(); it++){ 55 initializeTable(*it, dict); 56 visit[*it] = false; 57 } 58 return minLength(start, end, 1); 59 }
看了网上的一些分析，改成BFS，并且在最开始没有必要建图，那样复杂度是O（n^2）。因为每个单词每次只改变一个字母，所以可以对每个单词的每个字母进行修改，复杂度是O（26*lengthOfWord）。修改后的代码如下：
1 int ladderLength(string start, string end, unordered_set<string> &dict) { 2 // IMPORTANT: Please reset any member data you declared, as 3 // the same Solution instance will be reused for each test case. 4 dict.insert(end); 5 queue<pair<string, int> > q; 6 q.push(pair<string, int>(start, 1)); 7 int result = 0; 8 while(!q.empty()){ 9 string tmp = q.front().first; 10 int step = q.front().second; 11 if(tmp == end){ 12 result = step; 13 break; 14 } 15 q.pop(); 16 for(int i = 0; i < tmp.length(); i++){ 17 char tc = tmp[i]; 18 for(char c = 'a'; c <= 'z'; c++){ 19 if(tc == c) 20 continue; 21 tmp[i] = c; 22 if(dict.find(tmp) != dict.end()){ 23 q.push(pair<string, int>(tmp, step+1)); 24 dict.erase(tmp); 25 } 26 tmp[i] = tc; 27 } 28 } 29 } 30 return result; 31 }