[LeetCode] 953. Verifying an Alien Dictionary

Posted 2021-02-19 CNoodle

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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 lexicographically in this alien language.

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).

Constraints:

1 <= words.length <= 100
1 <= words[i].length <= 20
order.length == 26
All characters in words[i] and order are English lowercase letters.

验证外星语字典。题目即是题意，给了一些单词和一个定好的字典序，请你验证单词是否是有序的。这道题虽然看上去跟269很像但是会简单很多，思路是把order里面的字母之间的顺序转化为hashmap<letter, order>，然后再对words里面的单词进行两两比较。比较的方式是按字母逐个比较，如果同样位置上的两个字母不一样，则判断是否第一个单词的字母的字典序更小，若不是就return false；如果判断完了并没有找到false的情形，则判断是否第一个单词更短，因为除了满足字典序之外，也需要满足更短的单词在前。

时间O(n)

空间O(n)

Java实现

 1 class Solution {
 2     public boolean isAlienSorted(String[] words, String order) {
 3         HashMap<Character, Integer> map = new HashMap<>();
 4         for (int i = 0; i < order.length(); i++) {
 5             char ch = order.charAt(i);
 6             map.put(ch, i);
 7         }
 8 
 9         for (int i = 0; i < words.length - 1; i++) {
10             if (!inorder(words[i], words[i + 1], map)) {
11                 return false;
12             }
13         }
14         return true;
15     }
16 
17     private boolean inorder(String w1, String w2, HashMap<Character, Integer> map) {
18         for (int i = 0; i < w1.length() && i < w2.length(); i++) {
19             char ch1 = w1.charAt(i);
20             char ch2 = w2.charAt(i);
21             int index1 = map.get(ch1);
22             int index2 = map.get(ch2);
23             if (index1 < index2) {
24                 return true;
25             } else if (index1 > index2) {
26                 return false;
27             }
28         }
29         return w1.length() <= w2.length();
30     }
31 }