Java [Leetcode 357]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]

)

解题思路：

This problem can also be solved using a dynamic programming approach and some knowledge of combinatorics.

Let f(k) = count of numbers with unique digits with length equals k.

f(1) = 10, ..., f(k) = 9 * 9 * 8 * ... (9 - k + 2) [The first factor is 9 because a number cannot start with 0].

代码如下：

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