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 non-empty words from the dictionary, where words are sorted lexicographically by the rules of this new language. Derive the order of letters in this language.

Example 1:
Given the following words in dictionary,

[
  "wrt",
  "wrf",
  "er",
  "ett",
  "rftt"
]

The correct order is: "wertf".

Example 2:
Given the following words in dictionary,

[
  "z",
  "x"
]

The correct order is: "zx".

Example 3:
Given the following words in dictionary,

[
  "z",
  "x",
  "z"
]

The order is invalid, so return "".

Note:

1. You may assume all letters are in lowercase.
2. You may assume that if a is a prefix of b, then a must appear before b in the given dictionary.
3. If the order is invalid, return an empty string.
4. There may be multiple valid order of letters, return any one of them is fine.

一刷
題解：這題是根據(jù)具體的單詞找出字典序。
方法：先建立相鄰點(diǎn)的map<Character, Set<Character>>, 和出入度map<Character, Integer>, 為1度的點(diǎn)在前，移除要出，2度的點(diǎn)變?yōu)?度
由于不僅從左到右滿足alphabetic, 從上到下也滿足园蝠。
所以，上下相鄰的兩個(gè)word, 找到第一個(gè)不相同的首字母膀哲，那么一定是上面指向下面。（都只找第一個(gè)，因?yàn)楹竺娴捻樞虿淮_定）
上下相鄰的兩個(gè)字母堪嫂，找到第一個(gè)不一樣的首字母a, b，那么圖中一定是a->b. 如果采用word中相鄰字母的順序木柬，很容易形成環(huán)皆串。構(gòu)造了graph和degree之后再采用BFS

public String alienOrder(String[] words) {
    Map<Character, Set<Character>> map=new HashMap<Character, Set<Character>>();
    Map<Character, Integer> degree=new HashMap<Character, Integer>();
    String result="";
    if(words==null || words.length==0) return result;
    for(String s: words){
        for(char c: s.toCharArray()){
            degree.put(c,0);
        }
    }
    for(int i=0; i<words.length-1; i++){
        String cur=words[i];
        String next=words[i+1];
        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=new HashSet<Character>();
                if(map.containsKey(c1)) set=map.get(c1);
                if(!set.contains(c2)){
                    set.add(c2);
                    map.put(c1, set);
                    degree.put(c2, degree.get(c2)+1);
                }
                break;
            }
        }
    }
    Queue<Character> q=new LinkedList<Character>();
    for(char c: degree.keySet()){
        if(degree.get(c)==0) q.add(c);
    }
    while(!q.isEmpty()){
        char c=q.remove();
        result+=c;
        if(map.containsKey(c)){
            for(char c2: map.get(c)){
                degree.put(c2,degree.get(c2)-1);
                if(degree.get(c2)==0) q.add(c2);
            }
        }
    }
    if(result.length()!=degree.size()) return "";
    return result;
}