LeetCode 269. Alien Dictionary

原题链接在这里：https://leetcode.com/problems/alien-dictionary/

题目：

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.

题解：

采用BFS based topological sort. 建立graph 和 indegree array.

字典里有多个单词，这些竖着的单词是按照首字母排序的，如果首字母相同就看第二个字母，以此类推.

用queue把indegree为0的vertex加到queue中开始做BFS.

Note: when using while loop, first thing is to remember to increase index.

Time Complexity: O(V + E). Space: O(V). V最大26. Edge最大为words.length.

AC Java:

class Solution {
    public String alienOrder(String[] words) {
        if(words == null || words.length == 0){
            return "";
        }
        
        //看看words array中都含有哪些字母
        HashSet<Character> charSet = new HashSet<>();
        for(String w : words){
            for(char c : w.toCharArray()){
                charSet.add(c);
            }
        }
        
        //构建 adjancy list 形式的graph, 计算每个vertex 的indegree
        int [] in = new int[26];
        HashMap<Character, HashSet<Character>> graph = new HashMap<>();
        for(int i = 1; i < words.length; i++){
            String pre = words[i - 1];
            String cur = words[i];
            int j = 0;
            while(j < pre.length() && j < cur.length()){
                if(pre.charAt(j) != cur.charAt(j)){
                    char sour = pre.charAt(j);
                    char dest = cur.charAt(j);
                    
                    graph.putIfAbsent(sour, new HashSet<Character>());
                    if(!graph.get(sour).contains(dest)){
                        in[dest - 'a']++;
                    }
                    
                    graph.get(sour).add(dest);
                    break;
                }
                
                j++;
                if(j < pre.length() && j == cur.length()){
                    return "";
                }
            }
        }
        
        //BFS 形式的topologial sort
        StringBuilder sb = new StringBuilder();
        LinkedList<Character> que = new LinkedList<>();
        for(char c = 'a'; c <= 'z'; c++){
            if(in[c - 'a'] == 0 && charSet.contains(c)){
                que.add(c);
            }
        }
        
        while(!que.isEmpty()){
            char cur = que.poll();
            sb.append(cur);
            if(graph.containsKey(cur)){
                for(char c : graph.get(cur)){
                    in[c - 'a']--;
                    if(in[c - 'a'] == 0){
                        que.add(c);
                    }
                }
            }
        }
        
        //若是sb的length不等于uniqueChar的size, 说明剩下的部分有环
        return sb.length() == charSet.size() ? sb.toString() : "";
    }
}

类似Course Schedule.