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大神真多

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UTOO_NAIVE

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#我的方法:将数字转成字符串,然后循环遍历将第一个字符和最后一个字符比较

def is_palindrome(n): num = str(n) p = 0 flag = True while p < len(num) / 2: if num[p] != num[-p - 1]: flag = False break p = p + 1 return flag

#看评论的最佳答案:将数字转成字符串,将字符串用切片翻转,若是回数翻转后不变

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


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