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feat: add solutions to lc/lcof2 problems
lc No.0269 & lcof2 No.114.Alien Dictionary
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‎lcof2/剑指 Offer II 114. 外星文字典/README.md

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<!-- 这里可写通用的实现逻辑 -->
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用 g 数组记录在火星字典中的字母先后关系，`g[i][j] = true` 表示字母 `i + 'a'` 在字母 `j + 'a'` 的前面；用 s 数组记录当前字典出现过的字母，cnt 表示出现过的字母数。
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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 g 数组，这样遍历时间复杂度为 `O(n^3)`；但是我们发现其实比较相邻的两个单词就可以了，比如 `a < b < c` 则比较 `a < b` 和 `b < c`， a 和 c 的关系便确定了。因此算法可以优化成比较相邻两个单词，时间复杂度为 `O(n²)`。
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出现矛盾的情况：
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- `g[i][j] = g[j][i] = true`；
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- 后一个单词是前一个单词的前缀；
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- 在拓扑排序后 ans 的长度小于统计到的字母个数。
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拓扑排序：
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- 统计所有出现的字母入度；
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- 将所有入度为 0 的字母加入队列；
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- 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；
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- 得到的便是字母的拓扑序，也就是火星字典的字母顺序。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def alienOrder(self, words: List[str]) -> str:
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g = [[False] * 26 for _ in range(26)]
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s = [False] * 26
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cnt = 0
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n = len(words)
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for i in range(n - 1):
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for c in words[i]:
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if cnt == 26:
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break
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o = ord(c) - ord('a')
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if not s[o]:
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cnt += 1
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s[o] = True
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m = len(words[i])
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for j in range(m):
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if j >= len(words[i + 1]):
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return ''
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c1, c2 = words[i][j], words[i + 1][j]
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if c1 == c2:
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continue
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o1, o2 = ord(c1) - ord('a'), ord(c2) - ord('a')
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if g[o2][o1]:
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return ''
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g[o1][o2] = True
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break
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for c in words[n - 1]:
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if cnt == 26:
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break
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o = ord(c) - ord('a')
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if not s[o]:
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cnt += 1
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s[o] = True
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indegree = [0] * 26
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for i in range(26):
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for j in range(26):
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if i != j and s[i] and s[j] and g[i][j]:
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indegree[j] += 1
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q = deque()
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ans = []
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for i in range(26):
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if s[i] and indegree[i] == 0:
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q.append(i)
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while q:
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t = q.popleft()
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ans.append(chr(t + ord('a')))
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for i in range(26):
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if s[i] and i != t and g[t][i]:
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indegree[i] -= 1
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if indegree[i] == 0:
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q.append(i)
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return '' if len(ans) < cnt else ''.join(ans)
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public String alienOrder(String[] words) {
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boolean[][] g = new boolean[26][26];
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boolean[] s = new boolean[26];
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int cnt = 0;
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int n = words.length;
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for (int i = 0; i < n - 1; ++i) {
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for (char c : words[i].toCharArray()) {
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if (cnt == 26) {
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break;
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}
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c -= 'a';
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if (!s[c]) {
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++cnt;
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s[c] = true;
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}
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}
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int m = words[i].length();
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for (int j = 0; j < m; ++j) {
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if (j >= words[i + 1].length()) {
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return "";
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}
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char c1 = words[i].charAt(j), c2 = words[i + 1].charAt(j);
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if (c1 == c2) {
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continue;
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}
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if (g[c2 - 'a'][c1 - 'a']) {
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return "";
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}
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g[c1 - 'a'][c2 - 'a'] = true;
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break;
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}
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}
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for (char c : words[n - 1].toCharArray()) {
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if (cnt == 26) {
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break;
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}
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c -= 'a';
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if (!s[c]) {
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++cnt;
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s[c] = true;
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}
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}
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int[] indegree = new int[26];
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for (int i = 0; i < 26; ++i) {
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for (int j = 0; j < 26; ++j) {
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if (i != j && s[i] && s[j] && g[i][j]) {
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++indegree[j];
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}
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}
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}
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Deque<Integer> q = new LinkedList<>();
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for (int i = 0; i < 26; ++i) {
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if (s[i] && indegree[i] == 0) {
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q.offerLast(i);
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}
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}
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StringBuilder ans = new StringBuilder();
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while (!q.isEmpty()) {
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int t = q.pollFirst();
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ans.append((char) (t + 'a'));
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for (int i = 0; i < 26; ++i) {
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if (i != t && s[i] && g[t][i]) {
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if (--indegree[i] == 0) {
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q.offerLast(i);
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}
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}
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}
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}
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return ans.length() < cnt ? "" : ans.toString();
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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string alienOrder(vector<string>& words) {
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vector<vector<bool>> g(26, vector<bool>(26));
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vector<bool> s(26);
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int cnt = 0;
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int n = words.size();
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for (int i = 0; i < n - 1; ++i)
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{
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for (char c : words[i])
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{
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if (cnt == 26) break;
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c -= 'a';
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if (!s[c])
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{
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++cnt;
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s[c] = true;
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}
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}
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int m = words[i].size();
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for (int j = 0; j < m; ++j)
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{
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if (j >= words[i + 1].size()) return "";
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char c1 = words[i][j], c2 = words[i + 1][j];
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if (c1 == c2) continue;
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if (g[c2 - 'a'][c1 - 'a']) return "";
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g[c1 - 'a'][c2 - 'a'] = true;
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break;
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}
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}
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for (char c : words[n - 1])
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{
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if (cnt == 26) break;
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c -= 'a';
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if (!s[c])
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{
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++cnt;
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s[c] = true;
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}
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}
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vector<int> indegree(26);
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for (int i = 0; i < 26; ++i)
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for (int j = 0; j < 26; ++j)
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if (i != j && s[i] && s[j] && g[i][j])
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++indegree[j];
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queue<int> q;
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for (int i = 0; i < 26; ++i)
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if (s[i] && indegree[i] == 0)
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q.push(i);
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string ans = "";
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while (!q.empty())
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{
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int t = q.front();
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ans += (t + 'a');
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q.pop();
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for (int i = 0; i < 26; ++i)
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if (i != t && s[i] && g[t][i])
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if (--indegree[i] == 0)
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q.push(i);
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}
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return ans.size() < cnt ? "" : ans;
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}
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};
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```
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### **...**

‎lcof2/剑指 Offer II 114. 外星文字典/Solution.cpp

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class Solution {
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public:
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string alienOrder(vector<string>& words) {
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vector<vector<bool>> g(26, vector<bool>(26));
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vector<bool> s(26);
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int cnt = 0;
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int n = words.size();
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for (int i = 0; i < n - 1; ++i)
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{
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for (char c : words[i])
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{
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if (cnt == 26) break;
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c -= 'a';
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if (!s[c])
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{
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++cnt;
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s[c] = true;
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}
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}
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int m = words[i].size();
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for (int j = 0; j < m; ++j)
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{
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if (j >= words[i + 1].size()) return "";
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char c1 = words[i][j], c2 = words[i + 1][j];
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if (c1 == c2) continue;
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if (g[c2 - 'a'][c1 - 'a']) return "";
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g[c1 - 'a'][c2 - 'a'] = true;
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break;
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}
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}
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for (char c : words[n - 1])
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{
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if (cnt == 26) break;
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c -= 'a';
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if (!s[c])
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{
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++cnt;
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s[c] = true;
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}
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}
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vector<int> indegree(26);
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for (int i = 0; i < 26; ++i)
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for (int j = 0; j < 26; ++j)
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if (i != j && s[i] && s[j] && g[i][j])
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++indegree[j];
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queue<int> q;
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for (int i = 0; i < 26; ++i)
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if (s[i] && indegree[i] == 0)
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q.push(i);
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string ans = "";
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while (!q.empty())
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{
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int t = q.front();
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ans += (t + 'a');
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q.pop();
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for (int i = 0; i < 26; ++i)
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if (i != t && s[i] && g[t][i])
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if (--indegree[i] == 0)
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q.push(i);
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}
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return ans.size() < cnt ? "" : ans;
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}
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};

‎lcof2/剑指 Offer II 114. 外星文字典/Solution.java

Lines changed: 73 additions & 0 deletions
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class Solution {
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public String alienOrder(String[] words) {
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boolean[][] g = new boolean[26][26];
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boolean[] s = new boolean[26];
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int cnt = 0;
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int n = words.length;
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for (int i = 0; i < n - 1; ++i) {
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for (char c : words[i].toCharArray()) {
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if (cnt == 26) {
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break;
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}
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c -= 'a';
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if (!s[c]) {
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++cnt;
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s[c] = true;
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}
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}
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int m = words[i].length();
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for (int j = 0; j < m; ++j) {
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if (j >= words[i + 1].length()) {
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return "";
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}
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char c1 = words[i].charAt(j), c2 = words[i + 1].charAt(j);
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if (c1 == c2) {
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continue;
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}
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if (g[c2 - 'a'][c1 - 'a']) {
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return "";
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}
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g[c1 - 'a'][c2 - 'a'] = true;
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break;
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}
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}
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for (char c : words[n - 1].toCharArray()) {
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if (cnt == 26) {
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break;
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}
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c -= 'a';
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if (!s[c]) {
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++cnt;
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s[c] = true;
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}
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}
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int[] indegree = new int[26];
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for (int i = 0; i < 26; ++i) {
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for (int j = 0; j < 26; ++j) {
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if (i != j && s[i] && s[j] && g[i][j]) {
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++indegree[j];
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}
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}
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}
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Deque<Integer> q = new LinkedList<>();
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for (int i = 0; i < 26; ++i) {
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if (s[i] && indegree[i] == 0) {
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q.offerLast(i);
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}
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}
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StringBuilder ans = new StringBuilder();
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while (!q.isEmpty()) {
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int t = q.pollFirst();
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ans.append((char) (t + 'a'));
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for (int i = 0; i < 26; ++i) {
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if (i != t && s[i] && g[t][i]) {
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if (--indegree[i] == 0) {
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q.offerLast(i);
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}
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}
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}
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}
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return ans.length() < cnt ? "" : ans.toString();
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}
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}

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