超时了超时了。。但我觉得是对的

class Solution(object):
    def countNumbersWithUniqueDigits(self, n):
        """
        :type n: int
        :rtype: int
        """
        if  n < 1:
            return 1
        if  n == 1:
            return 10
        i = 1
        res = 0
        while i <= n:
            visit = [0 for j in range(10)]
            count = [0]
            self.backtrack(i, count, [], visit)
            res += count[0]
            i += 1
        return res
            
    def backtrack(self, num, count, temp, visit):
        if len(temp) == num:
            if temp[0] == 0 and num != 1:
                return
            else:
                count[0] += 1
                return
        for i in range(10):
            if visit[i] == 0:
                visit[i] = 1
                self.backtrack(num, count, temp+[i], visit)
                visit[i] = 0

一种用排列组合公式做的：

class Solution(object):
    def countNumbersWithUniqueDigits(self, n):
        """
        :type n: int
        :rtype: int
        """
        if  n < 1:
            return 1
        if  n == 1:
            return 10
        i = n 
        res = 0
        while i > 1:
            count = 9
            j = 9
            for m in range(i-1):
                count *= j
                j -= 1
            res += count
            i -= 1
        return res + 10

这一题的具体的一个说明：
This is a digit combination problem. Can be solved in at most 10 loops.

When n == 0, return 1. I got this answer from the test case.

When n == 1, _ can put 10 digit in the only position. [0, ... , 10]. Answer is 10.

When n == 2, _ _ first digit has 9 choices [1, ..., 9], second one has 9 choices excluding the already chosen one. So totally 9 * 9 = 81. answer should be 10 + 81 = 91

When n == 3, _ _ _ total choice is 9 * 9 * 8 = 684. answer is 10 + 81 + 648 = 739

When n == 4, _ _ _ _ total choice is 9 * 9 * 8 * 7.

...

When n == 10, _ _ _ _ _ _ _ _ _ _ total choice is 9 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1

When n == 11, _ _ _ _ _ _ _ _ _ _ _ total choice is 9 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 * 0 = 0

按照这种写法

class Solution(object):
    def countNumbersWithUniqueDigits(self, n):
        """
        :type n: int
        :rtype: int
        """
        if n == 0:
            return 1
        ans = 10
        base = 9
        for i in range(2, min(n+1, 11)):
            base = base *(9-i+2)
            ans += base
        return ans