如何使用方法缩短此代码

Question

#Exercise 9: Check Palindrome Number
#Numbers stored in x/y
x = 121
y = 125
#Transferred int in list
num_y = list(str(y))
num_x = list(str(x))
#Flipping the list
rev_y = num_y[::-1]
rev_x = num_x[::-1]
#Compare the flip if they are palindrome
if num_y == rev_y:
print("Yes. given number is palindrome number")
else:
print("No. given number is not palindrome number")
if num_x == rev_x:
print("Yes. given number is palindrome number")
else:
print("No. given number is not palindrome number")

Answer 1

使用def定义函数
不需要将整型参数从字符串转换为列表。只需对字符串进行操作-它支持切片反转字符串。
根据字符串是否与反向字符串相同，从函数返回True或False。

def is_palindrome(n):
s = str(n)
return s == s[::-1]
x = 121
print(is_palindrome(x))
# True
y = 125
print(is_palindrome(y))
# False

Answer 2

这是排序方法:

num = '1421'
if num == str(num)[::-1]:
print('The given number is PALINDROME')
else:
print('The given number is NOT a palindrome')

输出:

The given number is NOT a palindrome

这对你有帮助。

Answer 3

你应该使用函数。

def print_if_number_is_palindrom(number):
if str(number) == str(number)[::-1]:
print("Yes, given number is palindrome number")
else:
print("No, given number is not palindrome number")
x = 121
y = 125
print_if_number_is_palindrom(x)
print_if_number_is_palindrom(y)

Answer 4

目标:

• 清理代码
• 将主逻辑移动到函数

方法:

• 创建通用函数，可用于任何类型的回文
• 将数字作为字符串传入函数

def is_palindrome(term):
for i in range(len(term)):
if term[i] != term[(len(term)-i-1)]:
return False
return True

num_x = str(121)
print('num_x palendrome?',is_palindrome(num_x))

输出→"num_x回文呢?真正的'

num_y = str(122)
print('num_y palendrome?',is_palindrome(num_y))

输出→"num_x回文呢?假'

注意:

从技术上讲，方法是在类中定义的函数。因此，您实际要求的是:

class ClassExample():

def __init__(self):
self.string = ''

def is_palindrome(self,term):
for i in range(len(term)):
if term[i] != term[(len(term)-i-1)]:
return False
return True

ex = ClassExample()

测试:

ex.is_palindrome('121')