Python实现寻找回文数字过程解析

回文数字是很有意思的数字，不管从最高位开始念，还是从个位开始念，最终结果都一样，有一种对称美。

下面是回文数字的函数判断方式：

def is_palindrome(n):
  str_num = str(n)
  len_num = len(str_num)
  if len_num <= 2 and str_num[0] == str_num[-1]:
    return True
  else:
    half_len = round(len_num/2)
    for i in range(half_len):
      if not str_num[i] == str_num[-(i+1)]:
        return False
    return True

下面我们举个栗子，看3000以内有哪些回文数，以及有多少个这样的数字：

result = list(filter(is_palindrome, range(0, 3000)))
print(result, '\n', len(result))

输出如下：

[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, 44, 55, 66, 77, 88, 99, 101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, 212, 222, 232, 242, 252, 262, 272, 282, 292, 303, 313, 323, 333, 343, 353, 363, 373, 383, 393, 404, 414, 424, 434, 444, 454, 464, 474, 484, 494, 505, 515, 525, 535, 545, 555, 565, 575, 585, 595, 606, 616, 626, 636, 646, 656, 666, 676, 686, 696, 707, 717, 727, 737, 747, 757, 767, 777, 787, 797, 808, 818, 828, 838, 848, 858, 868, 878, 888, 898, 909, 919, 929, 939, 949, 959, 969, 979, 989, 999, 1001, 1111, 1221, 1331, 1441, 1551, 1661, 1771, 1881, 1991, 2002, 2112, 2222, 2332, 2442, 2552, 2662, 2772, 2882, 2992]
129

可以看到3000以内的自然数中，有129个回文数。