126. Word Ladder II(Leetcode每日一題-2020.06.07）

Problem

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.

Example1

Input:
beginWord = “hit”,
endWord = “cog”,
wordList = [“hot”,“dot”,“dog”,“lot”,“log”,“cog”]
Output:
[
[“hit”,“hot”,“dot”,“dog”,“cog”],
[“hit”,“hot”,“lot”,“log”,“cog”]
]

Example2

Input:
beginWord = “hit”
endWord = “cog”
wordList = [“hot”,“dot”,“dog”,“lot”,“log”]
Output: []
Explanation: The endWord “cog” is not in wordList, therefore no possible transformation.

Solution

const int INF = 1 << 20;

class Solution {
private:
    unordered_map<string, int> wordId;
    vector<string> idWord;
    vector<vector<int>> edges;
public:
    vector<vector<string>> findLadders(string beginWord, string endWord, vector<string>& wordList) {
        int id = 0;
        for (const string& word : wordList) {
            if (!wordId.count(word)) {
                wordId[word] = id++;
                idWord.push_back(word);
            }
        }
        if (!wordId.count(endWord)) {
            return {};
        }
        if (!wordId.count(beginWord)) {
            wordId[beginWord] = id++;
            idWord.push_back(beginWord);
        }
        edges.resize(idWord.size());
        for (int i = 0; i < idWord.size(); i++) {
            for (int j = i + 1; j < idWord.size(); j++) {
                if (transformCheck(idWord[i], idWord[j])) {
                    edges[i].push_back(j);
                    edges[j].push_back(i);
                }
            }
        }
        const int dest = wordId[endWord];
        vector<vector<string>> res;
        queue<vector<int>> q;
        vector<int> cost(id, INF);
        q.push(vector<int>{wordId[beginWord]});
        cost[wordId[beginWord]] = 0;
        while (!q.empty()) {
            vector<int> now = q.front();
            q.pop();
            int last = now.back();
            if (last == dest) {
                vector<string> tmp;
                for (int index : now) {
                    tmp.push_back(idWord[index]);
                }
                res.push_back(tmp);
            } else {
                for (int i = 0; i < edges[last].size(); i++) {
                    int to = edges[last][i];
                    if (cost[last] + 1 <= cost[to]) {
                        cost[to] = cost[last] + 1;
                        vector<int> tmp(now);
                        tmp.push_back(to);
                        q.push(tmp);
                    }
                }
            }
        }
        return res;
    }

    bool transformCheck(const string& str1, const string& str2) {
        int differences = 0;
        for (int i = 0; i < str1.size() && differences < 2; i++) {
            if (str1[i] != str2[i]) {
                ++differences;
            }
        }
        return differences == 1;
    }
};

//作者：LeetCode-Solution
//鏈接：https://leetcode-cn.com/problems/word-ladder-ii/solution/dan-ci-jie-long-ii-by-leetcode-solution/