Question

代码有点混乱，但重点是将其划分为不同的函数。

我知道您可以测试回文，但是如何使用布尔变量，而不是直接说明打印功能。

def trial():
    palindrome = True

    c = str(input("Hey: "))

    for i in range(len(c)):
        a = c[i]
        s = c[-(i+1)]
    if  (a == s):       
        return 
    break
    else:
        print ("No, its not a plaindrome")
    break

    return palindrom 

def output():
    print True


trial()
output()

Answer 1

您可以在其他功能中使用return值

def isPalindrome(s):
    return s == s[::-1]

def output(s):
    if isPalindrome(s):           # We use the return value of the above function
        print('it is a palindrome')
    else:
        print('not a palindrome')

例如

>>> output('hello')
not a palindrome

>>> output('abcba')
it is a palindrome

Answer 2

您的编程理念的一些背景 - 我称之为（也可能是其他人）见证。这是什么意思？这意味着我们假设somethnig是True - 直到我们被证明是错误的。在现代法庭上也是如此，除非被证明有罪，否则有人是无辜的。

现在让我们看看代码：

# let's go on to the trial!
def trial():
    # the string is a palindrome, or "innocent" (this is redundant in this case)
    palindrome = True
    c = str(input("Hey: "))

    # this iterates the string twice, you actually need half
    # but that's only an optimization.
    for i in range(len(c)):
        a = c[i]
        s = c[-(i+1)]
        # if we found characters that are not mirrored - this is not a palindrome.
        # these are our witnesses!
        if  (a != s):        
            return False # naive: palindrome = False
    # we've gone over the string and found no witnesses to say it's "guilty"
    return True # naive: return palindrome

请记住，在使用这种风格时，你只关心＆＃34;有罪＆＃34;部分，然后你可以立即return。

Answer 3

这是一种检查回文的简单方法：

def is_palindromic(s):
    return s == s[::-1]

然后使用if语句中的返回值。

if is_palindromic(s):
    print "Yay!"
else:
    print "Too bad"

Python回文编程

3 个答案: