Leetcode 953. Verifying an Alien Dictionary

Seawolf

1. 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.
   
2. 
   
3. 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.
   
4. 
   
5. 
   
6. 
   
7. Example 1:
   
8. 
   
9. Input: words = ["hello","leetcode"], order = "hlabcdefgijkmnopqrstuvwxyz"
   
10. Output: true
    
11. Explanation: As 'h' comes before 'l' in this language, then the sequence is sorted.
    
12. Example 2:
    
13. 
    
14. Input: words = ["word","world","row"], order = "worldabcefghijkmnpqstuvxyz"
    
15. Output: false
    
16. Explanation: As 'd' comes after 'l' in this language, then words[0] > words[1], hence the sequence is unsorted.
    
17. Example 3:
    
18. 
    
19. Input: words = ["apple","app"], order = "abcdefghijklmnopqrstuvwxyz"
    
20. Output: false
    
21. 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).
    
22. 
    
23. 
    
24. Constraints:
    
25. 
    
26. 1 <= words.length <= 100
    
27. 1 <= words[i].length <= 20
    
28. order.length == 26
    
29. All characters in words[i] and order are English lowercase letters.

复制代码

1. class Solution:
   
2.     def isAlienSorted(self, words: List[str], order: str) -> bool:
   
3.         for i in range(1, len(words)):
   
4.             first = words[i - 1]
   
5.             second = words[i]
   
6.             length = min(len(first), len(second))
   
7.             if length == len(second) and first[:length] == second:
   
8.                 return False
   
9.             
   
10.             j = 0
    
11.             while j < length and first[j] == second[j]:
    
12.                 j += 1
    
13.             if j >= len(first): return True
    
14.             index_first = order.find(first[j])
    
15.             index_second = order.find(second[j])
    
16.             
    
17.             if index_first > index_second:
    
18.                 return False
    
19.             
    
20.         return True

复制代码