题目描述
给定一组单词和字母顺序，然后判断单词之间是否按字典序顺序排序。
字典序的理解：
设想一本英语字典里的单词，哪个在前哪个在后？
显然的做法是先按照第一个字母、以 a、b、c……z 的顺序排列；如果第一个字母一样，那么比较第二个、第三个乃至后面的字母。如果比到最后两个单词不一样长（比如，sigh 和 sight），那么把短者排在前。
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).
思路
创建一个字典序，代表每个字母的大小。然后将单词的排序转换为数字字符串排序：
比如'00'<'01','22'<'23'等
代码实现
class Solution: def isAlienSorted(self, words: 'List[str]', order: 'str') -> 'bool': ld=dict() for i in range(25,-1,-1): if i<10: ld[order[25-i]]='0'+str(i) else: ld[order[25-i]]=str(i) # print(ld) for i in range(len(words)-1): lw,rw='','' for l in words[i]: lw=lw+str(ld[l]) for l in words[i+1]: rw=rw+str(ld[l]) lw=lw+(20-len(lw))*'9' rw=rw+(20-len(rw))*'9' # print(lw,rw) if lw<rw: return False return True
其他解法
自己写的运行效率还可以36ms，超过99.7%。然后去讨论区看了下大佬写的，顿时给跪了,思路差不多，但是简洁多了：
class Solution: def isAlienSorted(self, words, order): m = {c: i for i, c in enumerate(order)} words = [[m[c] for c in w] for w in words] return all(w1 <= w2 for w1, w2 in zip(words, words[1:]))
