题解 | #简单错误记录#
简单错误记录
http://www.nowcoder.com/practice/67df1d7889cf4c529576383c2e647c48
用两个字典来维护排序数。一个记录出现的次数，一个记录出现先后。
dict_num={}#用于维护行数数目
dict_appear={}#用于维护记录出现的次数
dict_order={}#用于维护总的排序树
n = 0
while 1:
try:
s = input().split('\\')[-1]
data = s.split(' ')[0][-16:] + ' ' + s.split(' ')[1]
if data not in dict_num:
dict_num[data] = 1
dict_appear[data] = n
else:
dict_num[data] += 1
n += 1
except:
break
for ele in dict_num:
dict_order[ele] = -1000*dict_num[ele]+dict_appear[ele]
dict_order = sorted(dict_order.items(), key = lambda x: x[1])
#print(dict_order)
#print(dict_num)
for n, ele in enumerate(dict_order):
if n==8:
break
print(ele[0].split()[0][-16:], ele[0].split()[1], dict_num[ele[0]])
# for i in range(len(l[-8:])):
# print(l[-8:][i], ll[-8:][i])