Count Numbers with Unique Digits

1问题描述
Given a non-negative integer n, count all numbers with unique digits, x, where 0 ≤ x < 10的n次方.

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])
2.思路.
利用排列组合来解决.
3代码

class Solution {public:   int permutation(int n, int r)    {        if(r == 0)        {            return 1;        }else{            return n * permutation(n - 1, r - 1);        }    }    int countNumbersWithUniqueDigits(int n) {        int sum = 1;        if(n > 0)        {           int end = (n > 10)? 10 : n;           for(int i = 0; i < end; i++)           {               sum += 9 * permutation(9, i);           }        }        return sum;    }};