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”]

• 牛客网上oj输出顺序
• 思路很清楚，和Word Ladder I类似，只是需要记录输出的路径！
• case通过率为47.22%
• 参考leetcode：https://leetcode.com/problems/word-ladder-ii/discuss/40452

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

• 126. Word Ladder II