【leetcode】953. Verifying an Alien Dictionary

题目如下：

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. 1 <= words.length <= 100
2. 1 <= words[i].length <= 20
3. order.length == 26
4. All characters in words[i] and order are english lowercase letters.

解题思路：题目很长，但是很简单，说的简单点就是自定义了一个字典序，用这个自定义的字典序判断数组中字符串是否是升序的。

代码如下：

class Solution(object):
    def isAlienSorted(self, words, order):
        """
        :type words: List[str]
        :type order: str
        :rtype: bool
        """
        dic = {}
        for i,v in enumerate(order):
            dic[v] = i
 
        def cmpf(s1,s2,dic):
            for i in range(min(len(s1),len(s2))):
                if dic[s1[i]] > dic[s2[i]]:
                    return 1
                elif dic[s1[i]] < dic[s2[i]]:
                    return -1
            if len(s1) > len(s2):
                return 1
            else:
                return -1
        for i in range(len(words)-1):
            if cmpf(words[i],words[i+1],dic) == 1:
                return False
        return True