有关于我
我的生活
这个网站
刷题之旅
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:
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. Solution: class Solution { public String alienOrder(String[] words) { Map<Character, Set<Character>> edges = new HashMap<>(); Map<Character, Set<Character>> inDegrees = new HashMap<>(); for (int i = 0; i < words.length; i ++) { String word = words[i]; for (char c : word.toCharArray()) { edges.put(c, edges.getOrDefault(c, new HashSet<>())); inDegrees.put(c, inDegrees.getOrDefault(c, new HashSet<>())); } if (i > 0) { String lastWord = words[i - 1]; if (word.length() < lastWord.length() && lastWord.startsWith(word)) { return ""; } int len = Math.min(lastWord.length(), word.length()); for (int k = 0; k < len; k ++) { if (lastWord.charAt(k) == word.charAt(k)) { continue; } char curr = word.charAt(k); Set<Character> parents = inDegrees.getOrDefault(curr, new HashSet<>()); parents.add(lastWord.charAt(k)); inDegrees.put(curr, parents); Set<Character> neis = edges.getOrDefault(lastWord.charAt(k), new HashSet<>()); neis.add(curr); edges.put(lastWord.charAt(k), neis); break; } } } StringBuilder sb = new StringBuilder(); Deque<Character> queue = new ArrayDeque<>(); Set<Character> visited = new HashSet<>(); for (Map.Entry<Character, Set<Character>> entry : inDegrees.entrySet()) { if (entry.getValue().size() == 0) { char c = entry.getKey(); queue.offer(c); visited.add(c); } } while (!queue.isEmpty()) { char c = queue.poll(); sb.append(c); for (char n : edges.get(c)) { inDegrees.get(n).remove(c); if (inDegrees.get(n).size() == 0 && visited.add(n)) { queue.offer(n); } } } if (sb.length() == edges.size()) { return sb.toString(); } return ""; } }
Please enable JavaScript to view the comments