126. Word Ladder II
题目
Given two words (beginWord and endWord), and a dictionary's word list, find all shortest transformation sequence(s) 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"]
Return
[
["hit","hot","dot","dog","cog"],
["hit","hot","lot","log","cog"]
]
Note:
Return an empty list 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.
解析
For the most voted solution, it is very complicated.
I do a BFS for each path
for example:
{hit} ->
{hit,hot} ->
{hit,hot,dot}/{hit,hot,lot} ->
[“hit”,“hot”,“dot”,“dog”]/[“hit”,“hot”,“lot”,“log”] ->
[“hit”,“hot”,“dot”,“dog”,“cog”],
[“hit”,“hot”,“lot”,“log”,“cog”]
class Solution_126 {
public:
vector<vector<string>> findLadders(string beginWord, string endWord, vector<string>& wordList) {
vector<vector<string>> res;
unordered_set<string> visit; //notice we need to clear visited word in list after finish this level of BFS
queue<vector<string>> q;
unordered_set<string> wordlist(wordList.begin(), wordList.end());
q.push({ beginWord });
bool flag = false; //to see if we find shortest path
while (!q.empty()){
int size = q.size();
for (int i = 0; i < size; i++){ //for this level
vector<string> cur = q.front();
q.pop();
vector<string> newadd = addWord(cur.back(), wordlist);
for (int j = 0; j < newadd.size(); j++){ //add a word into path
vector<string> newline(cur.begin(), cur.end());
newline.push_back(newadd[j]);
if (newadd[j] == endWord){
flag = true;
res.push_back(newline);
}
visit.insert(newadd[j]); // insert newadd word
q.push(newline);
}
}
if (flag)
break; //do not BFS further
for (auto it = visit.begin(); it != visit.end(); it++)
wordlist.erase(*it); //erase visited one
visit.clear();
}
sort(res.begin(),res.end());
return res;
}
// find words with one char different in dict
// hot->[dot,lot]
vector<string> addWord(string word, unordered_set<string>& wordlist){
vector<string> res;
for (int i = 0; i < word.size(); i++){
char s = word[i];
for (char c = 'a'; c <= 'z'; c++){
word[i] = c;
if (wordlist.count(word)) res.push_back(word);
}
word[i] = s;
}
return res;
}
};
题目来源