Given two words (start and end), and a dictionary, find all shortest transformation sequence(s) 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"]
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.
分析：
给定一个单词start, 问通过一系列变换是否能够得到单词end, 中间的单词都必须是给定的字典中的词。
其实就是 从最初的一个状态，能够找到一个path, 到达最终的一个状态。
Bingo! 图的搜索： 找到图中两个node之间的所有的最短路径。
图的搜索有 深度优先（DFS） 和 广度优先（BFS）两种。
BFS适用于解决寻找最短路径的问题，因此采用BFS
class Solution {
public:
vector<vector<string>> findLadders(string start, string end, unordered_set<string> &dict) {
//BFS each node in graph has multipul fathers
vector<vector<string> > result;
vector<string> path;
bool found = false;
unordered_set<string> visited;
unordered_map<string, vector<string> > father;
queue<string> current, next;
if(start.size() != end.size()) return result;
current.push(start);
while(!current.empty())
{
string str(current.front()); current.pop();
for(size_t i=0; i<str.size(); ++i)
{
string new_str(str);
for(char c = ‘a‘; c <= ‘z‘; ++c)
{
if(c == new_str[i]) continue;
swap(c, new_str[i]);
if(new_str == end || dict.count(new_str) > 0 && !visited.count(new_str)){
visited.insert(new_str);
next.push(new_str);
father[new_str].push_back(str);
}
swap(c, new_str[i]);
if(new_str == end){
found = true;
break;
}
}
}// end for
if(current.empty() && !found) swap(current, next);
}// end while
buildPath(father, path, start, end, result);
return result;
}
private:
void buildPath(unordered_map<string, vector<string> > &father, vector<string> &path,
const string &start, const string &word, vector<vector<string> > &result){
path.push_back(word);
if(word == start){
result.push_back(path);
reverse(result.back().begin(), result.back().end());
}
else{
for(auto f : father[word])
buildPath(father, path, start, f, result);
}
path.pop_back();
}
};