[英]How can i make this code shorter with a methond Python
#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")
你應該使用一個函數。
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)
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))
輸出 -> 'num_x 回文數? 真的'
num_y = str(122)
print('num_y palendrome?',is_palindrome(num_y))
輸出 -> 'num_x 回文數? 錯誤的'
從技術上講,方法是在類中定義的函數。 因此,您實際要求的是:
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()
測試:
ex.is_palindrome('121')
這是排序器方法:
num = '1421'
if num == str(num)[::-1]:
print('The given number is PALINDROME')
else:
print('The given number is NOT a palindrome')
輸出:
The given number is NOT a palindrome
這將對您有所幫助。
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