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])

一刷
題解：
f(k) = 9 * 9 * 8 * ... (9 - k + 2)

因?yàn)槿绻呀?jīng)有兩個(gè)數(shù)，那么接下來(lái)的選擇就是10-2,..., 以此類推。然后sum up f(k)

public class Solution {
    public int countNumbersWithUniqueDigits(int n) {
        if(n == 0) return 1;
        int res = 10, cnt = 9;
        for(int i=2; i<=n; i++){
            cnt *= (11-i);
            res += cnt;
        }
        return res;
    }
}

二刷
同上。

class Solution {
    public int countNumbersWithUniqueDigits(int n) {
        if(n == 0) return 1;
        int cnt = 9, res = 10;
        for(int i=2; i<=n; i++){
            cnt*= (11 - i);
            res += cnt;
        }
        return res;
    }
}