題目詳情
給定兩個單詞（beginWord 和 endWord）和一個字典 wordList，找出所有從 beginWord 到 endWord 的最短轉換序列。轉換需遵循如下規則：
每次轉換隻能改變一個字母。
轉換過程中的中間單詞必須是字典中的單詞。
說明:
- 如果不存在這樣的轉換序列，返回一個空列表。
- 所有單詞具有相同的長度。
- 所有單詞只由小寫字母組成。
- 字典中不存在重複的單詞。
- 你可以假設 beginWord 和 endWord 是非空的，且二者不相同。
示例 1:
輸入:
beginWord = "hit",
endWord = "cog",
wordList = ["hot","dot","dog","lot","log","cog"]
輸出:
[
["hit","hot","dot","dog","cog"],
["hit","hot","lot","log","cog"]
]
示例 2:
輸入:
beginWord = "hit"
endWord = "cog"
wordList = ["hot","dot","dog","lot","log"]
輸出: []
解釋: endWord "cog" 不在字典中，所以不存在符合要求的轉換序列。
——題目難度:困難
這道力扣困難級別的題目很難，借鑑了大神的思路和代碼，大致如下：
-下面代碼解題
class Solution {
public:
vector<vector<string>> findLadders(string beginWord, string endWord, vector<string>& wordList) {
vector<vector<string>> ans;
vector<string> path;
if(std::find(wordList.begin(), wordList.end(), endWord) == wordList.end()) return ans;
unordered_map<string,int> depth;
unordered_map<string,vector<string>> neighbor;
queue<string> que;
unordered_set<string> wordSet(wordList.begin(), wordList.end());
que.push(beginWord);
depth[beginWord] = 1;
while(!que.empty()) {
string cur = que.front();
que.pop();
int wordLen = cur.size();
for(int i=0;i<wordLen;i++)
{
string temp = cur;
for(char c = 'a'; c <= 'z'; c++)
{
temp[i] = c;
if(wordSet.count(temp)) {
if(depth.count(temp) == 0) {
depth[temp] = depth[cur] + 1;
que.push(temp);
neighbor[temp].push_back(cur);
} else if(depth[temp] == depth[cur] + 1) {
neighbor[temp].push_back(cur);
}
}
}
}
}
dfs(beginWord, endWord, path, neighbor, ans);
return ans;
}
void dfs(string &beginWord, string &cur, vector<string> path,
unordered_map<string,vector<string>> &neighbor, vector<vector<string>> &ans)
{
if(cur == beginWord) {
path.push_back(cur);
std::reverse(path.begin(), path.end());
ans.push_back(path);
return ;
}
path.push_back(cur);
for(string word: neighbor[cur])
{
dfs(beginWord, word, path, neighbor, ans);
}
}
};
結果