Leetcode/各大家 -- 269. Alien Dictionary(BFS)

269. Alien Dictionary（BFS）
Hard
https://leetcode.com/problems/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:

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

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思路： use map1: map to store <c, set of char after c>
           use map2: degree to store <c, # of char before c> //based on loop not the final one
           Iterate each two adjacent string, to fill map and update degree 
           use a queue to do BFS. add c to queue when its degree is 0. When remove a c, minus degree
           ! watch edge case,  loop case
Complexity: time: O(n) space: O(n

public class Solution {
    public String alienOrder(String[] words) {
        HashMap<Character, Set<Character>> map = new HashMap<Character, Set<Character>>();//<c, char after c>
        HashMap<Character, Integer> degree = new HashMap<Character, Integer>();//<c, # of char before c>
        StringBuilder res = new StringBuilder();
        //initialize degree map
        for(int i = 0; i < words.length; i++){
            char[] word = words[i].toCharArray();
            for(int j = 0; j < word.length; j++){
                degree.put(word[j], 0);
            }
        }
        //compare adjacent string & fill map
        for(int i = 0; i < words.length - 1; i++){
            String cur = words[i];
            String next = words[i + 1];
            int len = Math.min(cur.length(), next.length());
            for(int j = 0; j < len; j++){
                char c1 = cur.charAt(j);
                char c2 = next.charAt(j);
                if(c1 != c2){
                    Set<Character> set = new HashSet<Character>();//watch 'Set' declaration
                    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;//rest comparision is meaningless & not record it!
                }else{
                    //edge case - no order: ["wrtkj","wrt"] 1:next stop at end 2: cur still have lefts
                    if(j + 1 == next.length() && j + 1 < cur.length())
                        return "";
                }
            }
        }
        //BFS - use Queue to pop char in order
        Queue<Character> queue = new LinkedList<Character>();
        for(char c: degree.keySet()){
            if(degree.get(c)==0){
                queue.add(c);//eg:[zx,zy], c: z,x
            } 
        }
        while(!queue.isEmpty()){
            char cur = queue.remove();
            res.append(cur);
            if(map.containsKey(cur)){
                for(char c: map.get(cur)){
                    degree.put(c, degree.get(c) - 1);
                    if(degree.get(c) == 0)queue.add(c);//add next char
                }
            }
        }
        //avoid loops. only < possible -- eg: ["qd","ab"] res = qa
        if(res.length() != degree.size())return "";
        return res.toString();
    }
}

ex:   ["wrt","wrf","er","ett","rftt","a","t"] answer: weratf
                                   ?"f"why working
w0(degree) letter before
f1(t)
e1(w)
t1(r)2(a)
r1(e)
a1(r)