[leetcode]788. Rotated Digits @ python

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原题

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?

Example:
Input: 10
Output: 4
Explanation:
There are four good numbers in the range [1, 10] : 2, 5, 6, 9.
Note that 1 and 10 are not good numbers, since they remain unchanged after rotating.
Note:

N will be in range [1, 10000].

解法

遍历 + 字典.

代码

class Solution(object):
    def rotatedDigits(self, N):
        """
        :type N: int
        :rtype: int
        """
        def isValid(n):
            n_str = str(n)
            map = {'0':'0','1':'1','8':'8','2':'5','5':'2','6':'9','9':'6'}
            res = ''
            for d in n_str:
                if d not in map:
                    return False
                res += map[d]
            return res != n_str 
            
        count = 0
        for i in range(1, N+1):
            if isValid(i):
                count += 1
        return count