357. Count Numbers with Unique Digits

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])

代码如下：

 public class Solution {
     public int countNumbersWithUniqueDigits(int n) {
         if(n==0)
         return 1;
         int sum=0,p=0;

         if(n<=2)
         return (int)Math.pow(9,n)+countNumbersWithUniqueDigits(n-1);
         else{
             sum=countNumbersWithUniqueDigits(n-1);
             n=n-2;
             p=81;
             int q=8;
             while(n>0)
             {
                 p=p*q;
                 q--;
                 n--;
             }

         }
         return sum+p;
     }
 }