标签：from ems tran case path target class size max
Given two words (beginWord and endWord), and a dictionary‘s word list, find all shortest transformation sequence(s) from beginWord to endWord, such that:
Note:
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"] ]
Example 2:
Input: beginWord = "hit" endWord = "cog" wordList = ["hot","dot","dog","lot","log"] Output: [] Explanation: The endWord "cog" is not in wordList, therefore no possible transformation.
Approach #1: DFS. [C++]
class Solution {
public:
vector<vector<string>> findLadders(string beginWord, string endWord, vector<string>& wordList) {
unordered_set<string> visited;
unordered_set<string> wordList_(wordList.begin(), wordList.end());
vector<vector<string>> ans;
queue<vector<string>> paths;
paths.push({beginWord});
// record length of current shortest transformation sequence
int level = 1;
// record lenght of last conditional shortest transformation sequence
// if the target sequence‘s length shorter than minLevel we break the loop.
// int minLevel = INT_MAX;
while (!paths.empty()) {
vector<string> path = paths.front();
paths.pop();
if (path.size() > level) {
for (string s : visited) wordList_.erase(s);
visited.clear();
// if (path.size() > minLevel) {
// break;
// }
level = path.size();
}
string last = path.back();
for (int i = 0; i < last.length(); ++i) {
string last_ = last;
for (char c = ‘a‘; c <= ‘z‘; ++c) {
last_[i] = c;
if (wordList_.count(last_)) {
vector<string> newpath = path;
newpath.push_back(last_);
visited.insert(last_);
if (last_ == endWord) {
minLevel = level;
ans.push_back(newpath);
} else {
paths.push(newpath);
}
}
}
}
}
return ans;
}
};
Analysis:
https://leetcode.com/problems/word-ladder-ii/discuss/40434/C%2B%2B-solution-using-standard-BFS-method-no-DFS-or-backtracking
标签：from ems tran case path target class size max
原文地址：https://www.cnblogs.com/ruruozhenhao/p/10352397.html
