【LeetCode】953. Verifying an Alien Dictionary 解题报告（Python）-CFANZ编程社区

id： fuxuemingzhu

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题目描述

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.

题目大意

题目给出的所有字符都是小写字符，给了一个新的字母表顺序，问，给出的words数组，是不是有序的。

解题方法

直接依次进行判断即可。拿出两个相邻的字符串pre和after，然后判断他们的相同位置的每个字符的顺序，如果pre的某个位置小于after，说明这两个字符串是有序的，那么继续判断；如果Pre的某个位置大于after，说明不有序，直接返回False。这两部判断完成之后没结束，我们还要继续判断Example 3的情况，所以，需要判断pre的长度是不是大于after，并且after等于pre的前部分。

在遍历完所有的字符串之后都没有返回False，说明是有序的，那么返回True.

class Solution(object):
    def isAlienSorted(self, words, order):
        """
        :type words: List[str]
        :type order: str
        :rtype: bool
        """
        N = len(words)
        d = {c : i for i, c in enumerate(order)}
        for i in range(N - 1):
            pre, after = words[i], words[i + 1]
            if pre == after: continue
            _len = min(len(pre), len(after))
            for j in range(_len):
                if d[pre[j]] < d[after[j]]:
                    break
                elif d[pre[j]] > d[after[j]]:
                    return False
            if len(pre) > len(after) and pre[:_len] == after:
                return False
        return True