LeetCode-Count Numbers with Unique Digits

Given a non-negative integer 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

Constraints:

0 <= n <= 8

分析
F(0) = 1
F(1) = 10
F(2) = 10 +9(第一个数不能为0)9(可以从0~9中选出一个任意一个不与第一位相同的数)
F(3) = 10 +99 + 998
F(4) = …

F(n) = F(n-1) + 9 * 8 * …* 9-n+1;

    public int countNumbersWithUniqueDigits(int n) {
        if(n == 0) return 1;
        int sum = 10;
        int step = 9;
        int K = 9;
        for( int i = 0 ; i < n-1;i ++){
            step = step *(K -i);
            sum += step;
        }
        return sum;
    }