953 Verifying an Alien Dictionary 验证外星语词典
Description:
In an alien language, surprisingly they also use english lowercase letters, but possibly in a different order. The order of the alphabet is some permutation of lowercase letters.
Given a sequence of words written in the alien language, and the order of the alphabet, return true if and only if the given words are sorted lexicographicaly in this alien language.
Example:
Example 1:
Input: words = ["hello","leetcode"], order = "hlabcdefgijkmnopqrstuvwxyz"
Output: true
Explanation: As 'h' comes before 'l' in this language, then the sequence is sorted.
Example 2:
Input: words = ["word","world","row"], order = "worldabcefghijkmnpqstuvxyz"
Output: false
Explanation: As 'd' comes after 'l' in this language, then words[0] > words[1], hence the sequence is unsorted.
Example 3:
Input: words = ["apple","app"], order = "abcdefghijklmnopqrstuvwxyz"
Output: false
Explanation: The first three characters "app" match, and the second string is shorter (in size.) According to lexicographical rules "apple" > "app", because 'l' > '∅', where '∅' is defined as the blank character which is less than any other character (More info).
Note:
1 <= words.length <= 100
1 <= words[i].length <= 20
order.length == 26
All characters in words[i] and order are english lowercase letters.
题目描述:
某种外星语也使用英文小写字母，但可能顺序 order 不同。字母表的顺序（order）是一些小写字母的排列。
给定一组用外星语书写的单词 words，以及其字母表的顺序 order，只有当给定的单词在这种外星语中按字典序排列时，返回 true；否则，返回 false。
示例 :
示例 1：
输入：words = ["hello","leetcode"], order = "hlabcdefgijkmnopqrstuvwxyz"
输出：true
解释：在该语言的字母表中，'h' 位于 'l' 之前，所以单词序列是按字典序排列的。
示例 2：
输入：words = ["word","world","row"], order = "worldabcefghijkmnpqstuvxyz"
输出：false
解释：在该语言的字母表中，'d' 位于 'l' 之后，那么 words[0] > words[1]，因此单词序列不是按字典序排列的。
示例 3：
输入：words = ["apple","app"], order = "abcdefghijklmnopqrstuvwxyz"
输出：false
解释：当前三个字符 "app" 匹配时，第二个字符串相对短一些，然后根据词典编纂规则 "apple" > "app"，因为 'l' > '∅'，其中 '∅' 是空白字符，定义为比任何其他字符都小（更多信息）。
提示：
1 <= words.length <= 100
1 <= words[i].length <= 20
order.length == 26
在 words[i] 和 order 中的所有字符都是英文小写字母。
思路:
- 改写比较器的方法, 按照 order的顺序比较两个字符串的长度和对应位置的大小
- 按照 order的顺序重新给 words中的单词赋值, 再比较两个字符串的大小
时间复杂度O(mn), 空间复杂度O(1), 其中 n为 words数组的大小, m为数组中的每一个字符串的长度
代码:
C++:
class Solution
{
public:
bool isAlienSorted(vector<string>& words, string order)
{
int index[128] = {0};
for (int i = 0; i < 26; i++) index[order[i]] = i + 'a';
for (int i = 0; i < words.size(); i++) for (int j = 0; j < words[i].size(); j++) words[i][j] = index[words[i][j]];
for (int i = 1; i < words.size(); i++) if (words[i - 1] > words[i]) return false;
return true;
}
};
Java:
class Solution {
public boolean isAlienSorted(String[] words, String order) {
Comparator<String> comparator = (a, b) -> {
for (int i = 0; i < Math.min(a.length(), b.length()); i++) if (a.charAt(i) != b.charAt(i)) return order.indexOf(a.charAt(i)) - order.indexOf(b.charAt(i));
return a.length() - b.length();
};
for (int i = 1; i < words.length; i++) if (comparator.compare(words[i - 1], words[i]) > 0) return false;
return true;
}
}
Python:
class Solution:
def isAlienSorted(self, words: List[str], order: str) -> bool:
return words == sorted(words, key=lambda word: [order.index(x) for x in word])
