Verifying an Alien Dictionary

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 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.

分析：

把每个字母和它的位置对应，位置越靠前，字母越小。这样我们可以比较两个字符串的每个字母。如果不一样，就看字母的大小和字符串的大小排列是否一致。

 1 class Solution {
 2     public boolean isAlienSorted(String[] words, String order) {
 3         int[] mapping = new int[26];
 4         for (int i = 0; i < order.length(); i++) {
 5             mapping[order.charAt(i) - 'a'] = i;
 6         }
 7         for (int i = 1; i < words.length; i++) {
 8             if (bigger(words[i - 1], words[i], mapping)) {
 9                 return false;
10             }
11         }
12         return true;
13     }
14 
15     boolean bigger(String s1, String s2, int[] mapping) {
16         int n = s1.length(), m = s2.length();
17         for (int i = 0; i < n && i < m; ++i) {
18             if (s1.charAt(i) != s2.charAt(i)) {
19                 return mapping[s1.charAt(i) - 'a'] > mapping[s2.charAt(i) - 'a'];
20             }
21         }
22         return n > m;
23     }
24 }