class Solution(object):
def isValid(self, word):
if len(word)<3:
return False
countV,countC=0,0;#分别统计原音和辅音
for i in range(len(word)):
if word[i].isdigit() or word[i].isalpha():
if word[i]=="a" or word[i]=="e" or word[i]=="i" or word[i]=='o' or word[i]=='u' or word[i]=='A' or word[i]=='E' or word[i]=='I' or word[i]=='O' or word[i]=='U':
countC+=1
elif word[i].isalpha():
countV+=1
else:
return False
return countC>0 and countV>0
效果展示: 目录结构: alphabets.py
GENERAL {"simple": "%#*-:. ","complex": "$B%8&WM#*oahkbdpqwmZO0QLCJUYXzcvunxrjft/\|()1{}[]?-_~<>i!lI;:,\"^. "
}
# Full list could be found here…
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