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现有一种使用英语字母的外星文语言，这门语言的字母顺序与英语顺序不同。
给定一个字符串列表 words ，作为这门语言的词典，words 中的字符串已经 按这门新语言的字母顺序进行了排序 。
请你根据该词典还原出此语言中已知的字母顺序，并 按字母递增顺序 排列。若不存在合法字母顺序，返回 "" 。若存在多种可能的合法字母顺序，返回其中 任意一种 顺序即可。
字符串 s 字典顺序小于 字符串 t 有两种情况：
- 在第一个不同字母处，如果
s中的字母在这门外星语言的字母顺序中位于t中字母之前，那么s的字典顺序小于t。 - 如果前面
min(s.length, t.length)字母都相同，那么s.length < t.length时，s的字典顺序也小于t。
示例 1：
输入：words = ["wrt","wrf","er","ett","rftt"] 输出："wertf"
示例 2：
输入：words = ["z","x"] 输出："zx"
示例 3：
输入：words = ["z","x","z"] 输出："" 解释：不存在合法字母顺序，因此返回 "" 。
提示：
1 <= words.length <= 1001 <= words[i].length <= 100words[i]仅由小写英文字母组成
注意：本题与主站 269 题相同： https://leetcode.cn/problems/alien-dictionary/
用数组
一个很简单的想法是遍历每一个单词，比较该单词和其后的所有单词，把所有的先后关系更新进数组
出现矛盾的情况：
-
$g[i][j]$ =$g[j][i]$ =$true$ ； - 后一个单词是前一个单词的前缀；
- 在拓扑排序后
$ans$ 的长度小于统计到的字母个数。
拓扑排序：
- 统计所有出现的字母入度；
- 将所有入度为
$0$ 的字母加入队列； - 当队列不空，出队并更新其他字母的入度，入度为
$0$ 则入队，同时出队时将当前字母加入$ans$ 的结尾； - 得到的便是字母的拓扑序，也就是火星字典的字母顺序。
class Solution:
def alienOrder(self, words: List[str]) -> str:
g = [[False] * 26 for _ in range(26)]
s = [False] * 26
cnt = 0
n = len(words)
for i in range(n - 1):
for c in words[i]:
if cnt == 26:
break
o = ord(c) - ord('a')
if not s[o]:
cnt += 1
s[o] = True
m = len(words[i])
for j in range(m):
if j >= len(words[i + 1]):
return ''
c1, c2 = words[i][j], words[i + 1][j]
if c1 == c2:
continue
o1, o2 = ord(c1) - ord('a'), ord(c2) - ord('a')
if g[o2][o1]:
return ''
g[o1][o2] = True
break
for c in words[n - 1]:
if cnt == 26:
break
o = ord(c) - ord('a')
if not s[o]:
cnt += 1
s[o] = True
indegree = [0] * 26
for i in range(26):
for j in range(26):
if i != j and s[i] and s[j] and g[i][j]:
indegree[j] += 1
q = deque()
ans = []
for i in range(26):
if s[i] and indegree[i] == 0:
q.append(i)
while q:
t = q.popleft()
ans.append(chr(t + ord('a')))
for i in range(26):
if s[i] and i != t and g[t][i]:
indegree[i] -= 1
if indegree[i] == 0:
q.append(i)
return '' if len(ans) < cnt else ''.join(ans)class Solution {
public String alienOrder(String[] words) {
boolean[][] g = new boolean[26][26];
boolean[] s = new boolean[26];
int cnt = 0;
int n = words.length;
for (int i = 0; i < n - 1; ++i) {
for (char c : words[i].toCharArray()) {
if (cnt == 26) {
break;
}
c -= 'a';
if (!s[c]) {
++cnt;
s[c] = true;
}
}
int m = words[i].length();
for (int j = 0; j < m; ++j) {
if (j >= words[i + 1].length()) {
return "";
}
char c1 = words[i].charAt(j), c2 = words[i + 1].charAt(j);
if (c1 == c2) {
continue;
}
if (g[c2 - 'a'][c1 - 'a']) {
return "";
}
g[c1 - 'a'][c2 - 'a'] = true;
break;
}
}
for (char c : words[n - 1].toCharArray()) {
if (cnt == 26) {
break;
}
c -= 'a';
if (!s[c]) {
++cnt;
s[c] = true;
}
}
int[] indegree = new int[26];
for (int i = 0; i < 26; ++i) {
for (int j = 0; j < 26; ++j) {
if (i != j && s[i] && s[j] && g[i][j]) {
++indegree[j];
}
}
}
Deque<Integer> q = new LinkedList<>();
for (int i = 0; i < 26; ++i) {
if (s[i] && indegree[i] == 0) {
q.offerLast(i);
}
}
StringBuilder ans = new StringBuilder();
while (!q.isEmpty()) {
int t = q.pollFirst();
ans.append((char) (t + 'a'));
for (int i = 0; i < 26; ++i) {
if (i != t && s[i] && g[t][i]) {
if (--indegree[i] == 0) {
q.offerLast(i);
}
}
}
}
return ans.length() < cnt ? "" : ans.toString();
}
}class Solution {
public:
string alienOrder(vector<string>& words) {
vector<vector<bool>> g(26, vector<bool>(26));
vector<bool> s(26);
int cnt = 0;
int n = words.size();
for (int i = 0; i < n - 1; ++i) {
for (char c : words[i]) {
if (cnt == 26) break;
c -= 'a';
if (!s[c]) {
++cnt;
s[c] = true;
}
}
int m = words[i].size();
for (int j = 0; j < m; ++j) {
if (j >= words[i + 1].size()) return "";
char c1 = words[i][j], c2 = words[i + 1][j];
if (c1 == c2) continue;
if (g[c2 - 'a'][c1 - 'a']) return "";
g[c1 - 'a'][c2 - 'a'] = true;
break;
}
}
for (char c : words[n - 1]) {
if (cnt == 26) break;
c -= 'a';
if (!s[c]) {
++cnt;
s[c] = true;
}
}
vector<int> indegree(26);
for (int i = 0; i < 26; ++i)
for (int j = 0; j < 26; ++j)
if (i != j && s[i] && s[j] && g[i][j])
++indegree[j];
queue<int> q;
for (int i = 0; i < 26; ++i)
if (s[i] && indegree[i] == 0)
q.push(i);
string ans = "";
while (!q.empty()) {
int t = q.front();
ans += (t + 'a');
q.pop();
for (int i = 0; i < 26; ++i)
if (i != t && s[i] && g[t][i])
if (--indegree[i] == 0)
q.push(i);
}
return ans.size() < cnt ? "" : ans;
}
};func alienOrder(words []string) string {
n := len(words)
if n == 0 {
return ""
}
if n == 1 {
return words[0]
}
inDegree := make(map[byte]int)
graph := make(map[byte][]byte)
for _, word := range words {
for i := 0; i < len(word); i++ {
inDegree[word[i]] = 0
}
}
for i := 0; i < n-1; i++ {
w1, w2 := words[i], words[i+1]
minLen := len(w1)
if len(w2) < minLen {
minLen = len(w2)
}
foundDifference := false
for j := 0; j < minLen; j++ {
if w1[j] != w2[j] {
inDegree[w2[j]]++
graph[w1[j]] = append(graph[w1[j]], w2[j])
foundDifference = true
break
}
}
if !foundDifference && len(w1) > len(w2) {
return ""
}
}
queue := make([]byte, 0)
for k := range inDegree {
if inDegree[k] == 0 {
queue = append(queue, k)
}
}
res := make([]byte, 0)
for len(queue) > 0 {
node := queue[0]
queue = queue[1:]
res = append(res, node)
for _, next := range graph[node] {
inDegree[next]--
if inDegree[next] == 0 {
queue = append(queue, next)
}
}
}
if len(res) != len(inDegree) {
return ""
}
return string(res)
}class Solution {
func alienOrder(_ words: [String]) -> String {
var graph = Array(repeating: Set<Int>(), count: 26)
var indegree = Array(repeating: 0, count: 26)
var seen = Array(repeating: false, count: 26)
var letterCount = 0
for i in 0..<words.count - 1 {
for char in words[i] {
let index = Int(char.asciiValue! - Character("a").asciiValue!)
if !seen[index] {
seen[index] = true
letterCount += 1
}
}
let minLength = min(words[i].count, words[i + 1].count)
for j in 0..<minLength {
let char1 = words[i][words[i].index(words[i].startIndex, offsetBy: j)]
let char2 = words[i + 1][words[i + 1].index(words[i + 1].startIndex, offsetBy: j)]
if char1 != char2 {
let c1 = Int(char1.asciiValue! - Character("a").asciiValue!)
let c2 = Int(char2.asciiValue! - Character("a").asciiValue!)
if !graph[c1].contains(c2) {
graph[c1].insert(c2)
indegree[c2] += 1
}
break
}
if j == minLength - 1 && words[i].count > words[i + 1].count {
return ""
}
}
}
for char in words[words.count - 1] {
let index = Int(char.asciiValue! - Character("a").asciiValue!)
if !seen[index] {
seen[index] = true
letterCount += 1
}
}
var queue = [Int]()
for i in 0..<26 {
if seen[i] && indegree[i] == 0 {
queue.append(i)
}
}
var order = ""
while !queue.isEmpty {
let u = queue.removeFirst()
order += String(UnicodeScalar(u + Int(Character("a").asciiValue!))!)
for v in graph[u] {
indegree[v] -= 1
if indegree[v] == 0 {
queue.append(v)
}
}
}
return order.count == letterCount ? order : ""
}
}