一个基础面试题吧,感觉不太难,但在list那里栽了跟头,三个填空仅仅对了一个。至于下面的编程题,也是当时想到什么写什么的,所以没有去管是否存在简单的方法。
填空
关于list删除
a = [1,2,3,4]
for i in a:
del a[0]
print(a)
>>>[3,4]
a = [1,2,3,4]
for i in a:
a.remove(i)
print(a)
>>>[2,4]
a = [1,2,3,4]
for i in a:
a.pop()
print(a)
>>>[1,2]
写出下列代码运行结果
will = ['will',28,['python','html','javascript']]
wilber = will
will[0] = 'wilber'
will[2].append('css')
print(will)
print(wilber)
>>>['wilber', 28, ['python', 'html', 'javascript', 'css']]
>>>['wilber', 28, ['python', 'html', 'javascript', 'css']]
import copy
will = ['will',28,['python','html','javascript']]
wilber = copy.copy(will)
will[0] = 'wilber'
will[2].append('css')
print(will)
print(wilber)
>>>['wilber', 28, ['python', 'html', 'javascript', 'css']]
>>>['will', 28, ['python', 'html', 'javascript', 'css']]
import copy
will = ['will',28,['python','html','javascript']]
wilber = copy.deepcopy(will)
will[0] = 'wilber'
will[2].append('css')
print(will)
print(wilber)
>>>['wilber', 28, ['python', 'html', 'javascript', 'css']]
>>>['will', 28, ['python', 'html', 'javascript']]
代码
写个方法使得输出如下
输入:AwSfds
输出:A-Ww-Sss-Ffff-Ddddd-Ssssss
def accum(string):
strlist = list(string)
strpower = ''
strlower = ''
resultlist = []
for i in range(1,len(strlist)+1):
strpower = strlist[i-1].upper()
for j in range(i):
strlower = strlist[i-1].lower() * j
result = strpower + strlower
resultlist.append(result)
print("-".join(resultlist))
accum("AwSfds")
输入一个整数,输出比这个数大的下个整数
如:输入:2019 输出:2091
def next_num(num):
strlist = list(str(num))
for i in range(len(strlist)):
if len(strlist)-2 > 0:
if strlist[len(strlist)-1-i]> strlist[len(strlist)-2-i]:
temp = strlist[len(strlist)-1-i]
strlist[len(strlist)-1-i] = strlist[len(strlist)-2-i]
strlist[len(strlist)-2-i] = temp
break
else:
if strlist[len(strlist)-1]> strlist[len(strlist)-2]:
temp = strlist[len(strlist)-1-i]
strlist[len(strlist)-1] = strlist[len(strlist)-2]
strlist[len(strlist)-2] = temp
return int("".join(strlist))
next_num(366697864)
给出list a=[1,2,3,4],将a中偶数位置的元素加3,并求list的和
a = [1,2,3,4]
res = 0
for i in range(len(a)):
if i % 2 == 0 and i != 0:
a[i] += 3
res += a[i]
print(res)