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 public class Solution { 2     public int countNumbersWithUniqueDigits(int n) { 3          4         if (n < 1) return 1; 5         if (n == 1) return 10;  // 0 - 10 6         int count = countNumbersWithUniqueDigits(n - 1); // 1 到 n-1 位数。 7         int result = 9;  // there are 9 choices in the first number. 9 in the second number, 8 in the third, 7 in the fourth number....... 8         for (int i = 0; i < n - 1; i++) {  // n位数 9             result = result * (9 - i);10         }11         return count + result;12     }13 }