A transformation sequence from word beginWord to word endWord using a dictionary wordList is a sequence of words beginWord -> s1 -> s2 -> ... -> sk such that:
si for 1 <= i <= k is in wordList. Note that beginWord does not need to be in wordList.sk == endWordGiven two words, beginWord and endWord, and a dictionary wordList, return all the shortest transformation sequences from beginWord to endWord, or an empty list if no such sequence exists. Each sequence should be returned as a list of the words [beginWord, s1, s2, ..., sk].
Example 1:
Input: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"] Output: [["hit","hot","dot","dog","cog"],["hit","hot","lot","log","cog"]] Explanation: There are 2 shortest transformation sequences: "hit" -> "hot" -> "dot" -> "dog" -> "cog" "hit" -> "hot" -> "lot" -> "log" -> "cog"
Example 2:
Input: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log"] Output: [] Explanation: The endWord "cog" is not in wordList, therefore there is no valid transformation sequence.
Constraints:
1 <= beginWord.length <= 5endWord.length == beginWord.length1 <= wordList.length <= 500wordList[i].length == beginWord.lengthbeginWord, endWord, and wordList[i] consist of lowercase English letters.beginWord != endWordwordList are unique.class Solution {
public:
vector<vector<string>> findLadders(string beginWord, string endWord, vector<string>& wordList) {
vector<vector<string>> ans;
int n = size(wordList), src = -1, dst = -1;
for (int i = 0; i < n; i++) {
if (wordList[i] == beginWord) src = i;
else if (wordList[i] == endWord) dst = i;
}
if (dst == -1) return ans;
if (src == -1) {
wordList.push_back(beginWord);
src = n++;
}
vector<int> adj[505], parent[505], path = {dst};
for (int i = 0; i < n; i++) {
for (int j = i+1; j < n; j++) {
if (isAdj(wordList[i], wordList[j])) {
adj[i].push_back(j);
adj[j].push_back(i);
}
}
}
bfs(adj, parent, src);
dfs(wordList, ans, parent, path, dst);
return ans;
}
private:
bool isAdj(string& s1, string& s2) {
int dif = 0;
for (int i = 0; i < size(s1); i++)
dif += s1[i] != s2[i];
return dif == 1;
}
void bfs(vector<int> adj[], vector<int> parent[], int& src) {
int dist[505] = {};
fill(begin(dist), end(dist), 505);
dist[src] = 0;
queue<int> q;
q.push(src);
parent[src] = {-1};
while (!q.empty()) {
int v = q.front();
q.pop();
for (int u: adj[v]) {
if (dist[u] > dist[v] + 1) {
dist[u] = dist[v] + 1;
q.push(u);
parent[u] = {v};
} else if (dist[u] == dist[v] + 1)
parent[u].push_back(v);
}
}
}
void dfs(vector<string>& wordList, vector<vector<string>>& ans, vector<int> parent[], vector<int>& path, int v) {
if (v == -1) {
vector<string> tmp(size(path)-1);
transform(rbegin(path)+1, rend(path), begin(tmp), [&] (int& t) { return wordList[t]; });
ans.push_back(move(tmp));
return;
}
for (int u: parent[v]) {
path.push_back(u);
dfs(wordList, ans, parent, path, u);
path.pop_back();
}
}
};