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写了个回文字符串的判断

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杨炎泽

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def is_palindrome(n): i = 0 while i < len(n) / 2: #只循环一半 if n[i] != n[-i - 1]: #从头数第N个和从尾数第N个是否相等 return False #不想等返回false i += 1 return True #如果没有出现不相等的,返回true

print(list(filter(is_palindrome, ["ABA", "123321", "909", "22222a"])))


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