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写作业，不用[::-1]也挺简单

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用户7192832094

#1 Created at ... [Delete] [Delete and Lock User] 
def is_palindrome(n):
	s = str(n)
	for x in range(len(s)):
		return s[x] == s[len(s) - x - 1]

ddMelse

#2 Created at ... [Delete] [Delete and Lock User]

你的程序等价于

def is_palindrome(n):
    s = str(n)
    return s[0] == s[len(s) - 1]

ZERO星痕

#3 Created at ... [Delete] [Delete and Lock User]

def is_palindrome(n):

    s = str(n)

    flag = True

    for x in range(len(s)):

        if s[x] != s[len(s) - x - 1]:

            flag = False

            break

    return flag

江南烟雨人7411

#4 Created at ... [Delete] [Delete and Lock User]

反对 ddMelse。 用你这段代码，一万以内的数， 返回这些：9919, 9929, 9939, 9949, 9959, 9969, 9979, 9989。

爱上树的椰子蟹

#5 Created at ... [Delete] [Delete and Lock User]

二楼的写法好像是有些问题。只筛选出了第一位和倒数第一位相同的数字，这种写法只是在3位数以内的数字内可以做判断，超过3位数就会出问题了

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