# 953. Verifying an Alien Dictionary

**<font color=red>难度: Easy</font>**

## 刷题内容

> 原题连接

* https://leetcode.com/problems/verifying-an-alien-dictionary/

> 内容描述

```
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).
 

Note:

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

## 解题方案

> 思路 1
******- 时间复杂度: O(N * W)******- 空间复杂度: O(1)******

sb题没什么好说的

```python
class Solution:
    def isAlienSorted(self, words, order):
        """
        :type words: List[str]
        :type order: str
        :rtype: bool
        """
        lookup = {}
        idx = 0
        for c in order:
            lookup[c] = idx
            idx += 1
            
        def isValid(word1, word2):
            for i in range(min(len(word1), len(word2))):
                if lookup[word1[i]] > lookup[word2[i]]:
                    return False
                elif lookup[word1[i]] == lookup[word2[i]]:
                    continue
                else:
                    return True
            return len(word1) < len(word2)
        
        for i in range(len(words)-1):
            if not isValid(words[i], words[i+1]):
                return False
        return True
```




























