Count Numbers with Unique Digits

题目描述：

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]

)

思路是：恰好n位数形成无重复数是9*9*8*7....所以求出n位数构成的数，加上countNumbersWithUniqueDigits(n-1)即可。

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