求回数

Posted 2020-10-04 绝望的老猫

代码比较简单，重点是理解切片。

1 #-*- coding:utf-8 -*-
2 
3 def is_palindrome(n):
4     return str(n) == str(n)[::-1]
5 
6 output = filter(is_palindrome,range(1,1000))
7 print(list(output))

结果：

1 [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]

 

切片

[begin:end:step]

步长为-1时，然后翻转了整个字符串。