python -面试题--算法

44、编写程序，查找文本文件中最长的单词
def longest_word(filename):
with open(filename, ‘r’) as infile:
words = infile.read().split()
max_len = len(max(words, key=len))
return [word for word in words if len(word) == max_len]
print(longest_word(‘test.txt’))
[‘comprehensions’]
▍45、编写程序，检查序列是否为回文
a = input(“Enter The sequence”)
ispalindrome = a == a[::-1]
ispalindrome
True
▍46、编写程序，打印斐波那契数列的前十项
fibo = [0,1]
[fibo.append(fibo[-2]+fibo[-1]) for i in range(8)]
fibo
[0, 1, 1, 2, 3, 5, 8, 13, 21, 34]
▍47、编写程序，计算文件中单词的出现频率
from collections import Counter
def word_count(fname):
with open(fname) as f:
return Counter(f.read().split())
print(word_count(“test.txt”))
▍48、编写程序，输出给定序列中的所有质数
lower = int(input(“Enter the lower range:”))
upper = int(input(“Enter the upper range:”))
list(filter(lambda x:all(x % y != 0 for y in range(2, x)), range(lower, upper)))
Enter the lower range:10
Enter the upper range:50
[11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47]
▍49、编写程序，检查数字是否为Armstrong
图片
将每个数字依次分离，并累加其立方(位数)。
最后，如果发现总和等于原始数，则称为阿姆斯特朗数(Armstrong)。
num = int(input(“Enter the number:\n”))
order = len(str(num))
sum = 0
temp = num
while temp > 0:
digit = temp % 10
sum += digit ** order
temp //= 10
if num == sum:
print(num,“is an Armstrong number”)
else:
print(num,“is not an Armstrong number”)
▍50、用一行Python代码，从给定列表中取出所有的偶数和奇数
a = [1,2,3,4,5,6,7,8,9,10]
odd, even = [el for el in a if el % 2 == 1], [el for el in a if el % 2 ==0]
print(odd,even)
([1, 3, 5, 7, 9], [2, 4, 6, 8, 10])