leetcode357: Count Numbers with Unique Digits

要求：

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]

)

注意：判断n=0的情况

n=1 return 10

n=2 return 9*9+10

n=3 return 9*9*8+9*9+10

...

public static int countNumbersWithUniqueDigits(int n) {
if (n == 0)
return 1;
int flag = n;
int sum = 0;
for (int i = 0; i < n - 1; i++) {
int count = 9;
int pr = 9;
while (count > (10 - flag))
pr = pr * count--;
flag--; // 控制参与乘积的数字个数
sum += pr;
}
return sum + 10;
}