Alien Dictionary

There is a new alien language which uses the latin alphabet. However, the order among letters are unknown to you. You receive a list of words from the dictionary, where words are sorted lexicographically by the rules of this new language. Derive the order of letters in this language.

For example,
Given the following words in dictionary,

[
  "wrt",
  "wrf",
  "er",
  "ett",
  "rftt"
]
The correct order is: "wertf".

Note:
You may assume all letters are in lowercase.
If the order is invalid, return an empty string.
There may be multiple valid order of letters, return any one of them is fine.

Topology Sort, BFS

    public String alienOrder(String[] words) {

        if(words == null || words.length == 0) 
            return "";
        Map<Character, Set<Character>> map = new HashMap<Character, Set<Character>>();
        Map<Character, Integer> degree = new HashMap<Character, Integer>();
        StringBuilder sb = new StringBuilder();

        // put all word in-degree 0
        for(String s: words){
            for(char c: s.toCharArray()){
                degree.put(c,0);
            }
        }

        // compare each word and its pre-word char by char, 
        // if different, since c1 is in front of c2, put c2 into c1's next set, c2 in-dgree + 1
        for(int i=0; i < words.length-1; i++){
            String cur = words[i];
            String next = words[i+1];
            // using longer one
            int length = Math.min(cur.length(), next.length());
            for(int j=0; j<length; j++){
                char c1=cur.charAt(j);
                char c2=next.charAt(j);
                if(c1!=c2){
                    Set<Character> set = map.getOrDefault(c1, new HashSet());
                    if(!set.contains(c2)){
                        set.add(c2);
                        map.put(c1, set);
                        degree.put(c2, degree.get(c2) + 1);
                    }
                    break;
                }
            }
        }

        // topological sort via BFS
        Queue<Character> q = new LinkedList();
        // put all 0 in-degree into queue
        for(char c: degree.keySet()){
            if(degree.get(c) == 0) q.add(c);
        }
        while(!q.isEmpty()){
            char c = q.poll();
            sb.append(c);
            if(map.containsKey(c)){
                for(char c2: map.get(c)){
                    // all next chars' in-degree abstract 1
                    degree.put(c2,degree.get(c2) - 1);
                    if(degree.get(c2) == 0) q.add(c2);
                }
            }
        }
        if(sb.length() != degree.size()) return "";

        return sb.toString();
    }