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翻转法，囧

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千年C

#1 Created at ... [Delete] [Delete and Lock User] 
def is_palindrome(n):
    sn = str(n)
    if len(sn) % 2 == 0:
        return False
    else:
        k = 0
        while k<(len(sn)-1)/2:
            if sn[k]!=sn[-(k+1)]:
                return False
            k += 1
        return True

惫懒之

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

你这不对，回文数字是偶的也可以，比如22应该也算回文

fiinder

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

def is_palindrome(n): strn = str(n) for i in range(len(strn)//2): if int(strn[i]) != int(strn[-1 - i]): return False return True

output = filter(is_palindrome,range(1, 1000)) print(list(output))

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