题目要求
X is a good number if after rotating each digit individually by 180 degrees, we get a valid number that is different from X. Each digit must be rotated - we cannot choose to leave it alone.
A number is valid if each digit remains a digit after rotation. 0, 1, and 8 rotate to themselves; 2 and 5 rotate to each other; 6 and 9 rotate to each other, and the rest of the numbers do not rotate to any other number and become invalid.
Now given a positive number N, how many numbers X from 1 to N are good?
题目分析及思路
给定一个正整数范围，要求得到good number的个数。good number的定义是：对该数的各位数进行180度翻转，最后得到的数字与原数不同。这里，0，1，8翻转后还是本身，2，5，6，9翻转后可得不同的有效数字，剩余数字翻转则无效。可以先遍历这个范围的每一个数，并获得该数的各位数，最后只要确定各位数中没有3，4，7且有2，5，6，9即可确定该数为good number。
python代码
class Solution:
def rotatedDigits(self, N: int) -> int:
count = 0
for n in range(2,N+1):
digits = []
while n:
digit = n % 10
digits.append(digit)
n = n // 10
if not (set(digits) & set([3,4,7])) and (set(digits) & set([2,5,6,9])):
count += 1
return count