126. Word Ladder II

最新推荐文章于 2025-09-10 15:16:07 发布

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最新推荐文章于 2025-09-10 15:16:07 发布

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文章标签： c/c++

原文链接：http://www.cnblogs.com/ruruozhenhao/p/10352397.html

本文介绍了一种使用字典辅助的算法来解决单词阶梯II问题，即寻找从开始单词到目标单词的最短转换路径，每次仅改变一个字母，并确保每一步都在给定的字典中。通过深度优先搜索(DFS)的方法，实现了从开始单词到目标单词的所有最短路径的查找。

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.

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.

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

转载于:https://www.cnblogs.com/ruruozhenhao/p/10352397.html