滴滴数分笔试

# 第一题

SELECT t.uid,

CASE  WHEN A.level is null THEN 0 ELSE A.level END AS level,

CASE  WHEN B.mileage is null THEN 0 ELSE B.mileage END AS mileage,

CASE  WHEN C.complaint_cnt is null THEN 0 ELSE C.complaint_cnt END AS complaint_cnt

from

(select uid from A union select uid from B union select uid from C)t

left join A on t.uid=A.uid

left join B on t.uid = B.uid

left join C on t.uid=C.uid;

# 第二题

import collections
s=input().split(',')

count=collections.defaultdict(int)
str1=[]
# 初始化
for i in set(s):
count[str(i)]=0
for i in s:
count[str(i)]+=1
if count[str(i)]==1:
str1.append(str(i))
else:
str1.append(str(str(i)+'_'+str((count[str(i)]-2))))

print(str1)

编程题开始的时候只要25min 以为自己要写不完了，还好还好。。。