以下代码全部来自discuss。
1.dijkstra
https://leetcode.com/discuss/50930/java-solution-using-dijkstras-algorithm-with-explanation

public int ladderLength(String beginWord, String endWord, Set<String> wordDict) {
    Set<String> reached = new HashSet<String>();
    reached.add(beginWord);
    wordDict.add(endWord);
    int distance = 1;
    while(!reached.contains(endWord)) {
        Set<String> toAdd = new HashSet<String>();
        for(String each : reached) {
            for (int i = 0; i < each.length(); i++) {
                char[] chars = each.toCharArray();
                for (char ch = 'a'; ch <= 'z'; ch++) {
                    chars[i] = ch;
                    String word = new String(chars);
                    if(wordDict.contains(word)) {
                        toAdd.add(word);
                        wordDict.remove(word);
                    }
                }
            }
        }
        distance ++;
        if(toAdd.size() == 0) return 0;
        reached = toAdd;
    }
    return distance;
}

2.bfs
https://leetcode.com/discuss/42006/easy-76ms-c-solution-using-bfs

class Solution {
public:
    int ladderLength(string beginWord, string endWord, unordered_set<string>& wordDict) {
        wordDict.insert(endWord);
        queue<string> toVisit;
        addNextWords(beginWord, wordDict, toVisit);
        int dist = 2;
        while (!toVisit.empty()) {
            int num = toVisit.size();
            for (int i = 0; i < num; i++) {
                string word = toVisit.front();
                toVisit.pop();
                if (word == endWord) return dist;
                addNextWords(word, wordDict, toVisit);
            }
            dist++;
        }
    }
private:
    void addNextWords(string word, unordered_set<string>& wordDict, queue<string>& toVisit) {
        wordDict.erase(word);
        for (int p = 0; p < (int)word.length(); p++) {
            char letter = word[p];
            for (int k = 0; k < 26; k++) { 
                word[p] = 'a' + k;
                if (wordDict.find(word) != wordDict.end()) {
                    toVisit.push(word);
                    wordDict.erase(word);
                }
            }
            word[p] = letter;
        } 
    } 
};

3.two-end bfs
https://leetcode.com/discuss/28573/share-my-two-end-bfs-in-c-80ms

//BFS， two-end method
//traverse the path simultaneously from start node and end node, and merge in the middle
//the speed will increase (logN/2)^2 times compared with one-end method
int ladderLength(string start, string end, unordered_set<string> &dict) {
    unordered_set<string> begSet, endSet, *set1, *set2;
    begSet.insert(start);
    endSet.insert(end);
    int h=1, K=start.size();
    while(!begSet.empty()&&!endSet.empty()){
        if(begSet.size()<=endSet.size()){   //Make the size of two sets close for optimization
            set1=&begSet;   //set1 is the forward set
            set2=&endSet;   //set2 provides the target node for set1 to search
        }
        else{
            set1=&endSet;
            set2=&begSet;
        }
        unordered_set<string> itmSet;   //intermediate Set
        h++;
        for(auto i=set1->begin();i!=set1->end();i++){
            string cur=*i;
            for(int k=0;k<K;k++){   //iterate the characters in string cur
                char temp=cur[k];
                for(int l=0;l<26;l++){  //try all 26 alphabets
                    cur[k]='a'+l;
                    auto f=set2->find(cur);
                    if(f!=set2->end())return h;
                    f=dict.find(cur);
                    if(f!=dict.end()){
                        itmSet.insert(cur);
                        dict.erase(f);
                    }
                }
                cur[k]=temp;
            }
        }
        swap(*set1, itmSet);
    }
    return 0;
}