题目描述：

LeetCode 357. Count Numbers with Unique Digits

Given a non-negative integer n, count all numbers with unique digits, x, where 0 ≤ x < 10ⁿ.

Example:

Given n = 2, return 91. (The answer should be the total numbers in the range of 0 ≤ x < 100, excluding [11,22,33,44,55,66,77,88,99])

1. A direct way is to use the backtracking approach.
2. Backtracking should contains three states which are (the current number, number of steps to get that number and a bitmask which represent which number is marked as visited so far in the current number). Start with state (0,0,0) and count all valid number till we reach number of steps equals to 10ⁿ.
3. This problem can also be solved using a dynamic programming approach and some knowledge of combinatorics.
4. Let f(k) = count of numbers with unique digits with length equals k.
5. f(1) = 10, ..., f(k) = 9 * 9 * 8 * ... (9 - k + 2) [The first factor is 9 because a number cannot start with 0].

题目大意：

给定一个非负整数n，计算所有各位数字均不同的数字x的个数，其中0 ≤ x < 10^n。

测试用例如题目描述。

提示：

1. 一个直接的办法是使用回溯法。
2. 回溯应当包含三个状态（当前数字，得到该数字需要的步数，以及一个比特掩码用来表示当前已经访问过哪些位）。从状态（0,0,0）开始，并计算直到10^n为止的有效数字个数。
3. 这个问题也可以使用动态规划方法与一些组合的知识来求解。
4. f(k) = 长度为k的不重复数位的数字个数。
5. f(1) = 10, ..., f(k) = 9 * 9 * 8 * ... (9 - k + 2) [首个因数是9，因为数字不能从0开始].

解题思路：

参考提示的动态规划与排列组合解法。

Python代码：

class Solution(object):
    def countNumbersWithUniqueDigits(self, n):
        """
        :type n: int
        :rtype: int
        """
        nums = [9]
        for x in range(9, 0, -1):
            nums += nums[-1] * x,
        return sum(nums[:n]) + 1

 

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