LeetCode [269] 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 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:

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

Output: "wertf"

Example 2:

Input:
[
  "z",
  "x"
]

Output: "zx"

Example 3:

Input:
[
  "z",
  "x",
  "z"
] 

Output: "" 

Explanation: 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.

class Solution {
    Map<Character, Set<Character>> links = new HashMap<>();
    public String alienOrder(String[] words) {
        int i = 0;
        boolean cont = true;

        while(cont)
        {
            String prefix0 = "";
            char pChar = '.';
            cont = false;
            for(String w : words)
            {
                if(w.length() < i) continue; //"er**", if i = 3 skip
                String prefix = w.substring(0, i);
                if(w.length() >= i+1)
                {
                    cont = true;
                    char c =  w.charAt(i);
                    if(!links.containsKey(c)) links.put(c, new HashSet<>());
                    if(prefix.equals(prefix0) && pChar!='.' && pChar != c)
                        links.get(pChar).add(c);
                    pChar = c;
                }else{//w.length() == i
                    // for case "abc","ab"
                    if(prefix.equals(prefix0) && pChar != '.') return "";
                }
                prefix0 = prefix;
            }
            i++;
        }

        Set<Character> visited = new HashSet<Character>();
        Set<Character> path = new HashSet<Character>();
        List<Character> seq = new ArrayList<Character>();

        for(Map.Entry e : links.entrySet())
        {
            char node = (char) e.getKey();
            if(!visited.contains(node))
            {
                path.add(node);
                if(hasCycle(visited, path, seq, node)) return "";
                path.remove(node);
            }
        }

        Collections.reverse(seq);
        return seq.stream().map(Object::toString).collect(Collectors.joining(""));
    }

    boolean hasCycle(Set<Character> visited, Set<Character> path, List<Character> seq, char c)
    {
        visited.add(c);
        for(char next : links.get(c))
        {
            if(path.contains(next)) return true;
            if(visited.contains(next)) continue;
            path.add(next);
            if(links.containsKey(next) && hasCycle(visited, path, seq, next)) return true;
            path.remove(next);
        }
        seq.add(c);
        return false;
    }
}

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class Solution { Map<Character, Set<Character>> links = new HashMap<>(); public String alienOrder(String[] words) { int len = words.length; int j = 0; boolean cont = true; while(cont){ String prefix0 = ""; char ch0 = '.'; cont = false; for(int i=0; i<len; ++i){ String w = words[i]; String prefix = words[i].substring(0, Math.min(w.length(), j)); if(j>=w.length()){ if(prefix.equals(prefix0) && ch0!='.') return ""; prefix0 = prefix; ch0 = '.'; }else{ cont = true; char ch = words[i].charAt(j); if(!links.containsKey(ch)) links.put(ch, new HashSet<>()); if(prefix.equals(prefix0) && ch0!=ch && ch0!='.'){ links.get(ch0).add(ch); } prefix0 = prefix; ch0 = ch; } } j++; } Set<Character> path = new HashSet<>(); Set<Character> visited = new HashSet<>(); List<Character> list = new ArrayList<>(); for(char node : links.keySet()){ if(visited.contains(node)) continue; path.add(node); if(hasCycle(path, visited, list, node)) return ""; path.remove(node); } String ret = ""; for(char c : list) ret += c; return ret; } boolean hasCycle(Set<Character> path, Set<Character> visited, List<Character> list, char node){ visited.add(node); for(char nxt : links.get(node)){ if(path.contains(nxt)) return true; if(visited.contains(nxt)) continue; path.add(nxt); if(hasCycle(path, visited, list, nxt)) return true; path.remove(nxt); } list.add(0, node); return false; } }