Count Numbers with Unique Digits

Given anon-negativeinteger n, count all numbers with unique digits, x, where 0 ≤ x < 10n.

Example:

Input: 
2
Output: 
91 

Explanation: 
The answer should be the total numbers in the range of 0 ≤ x <100, 
excluding 11,22,33,44,55,66,77,88,99

分析

就是从高位起loop每个位，每个往后的数，可选数字都只能少一个，排列组合思想

class Solution:
    def countNumbersWithUniqueDigits(self, n: int) -> int:
        if n == 0:
            return 1
        res = 10
        avail = uniq = 9
        for i in range(2,n+1):
            uniq = avail*uniq
            res+= uniq
            avail -=1
        return res