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).
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
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
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:]))