LeetCode Count Numbers with Unique Digits

原题链接在这里：https://leetcode.com/problems/count-numbers-with-unique-digits/description/

题目：

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])

题解：

f(n)代表n位时有多少number是全用的unique digit的.

f(1) = 10. (0,1,2,3,4,5,6,7,8,9)

f(2) = 9*9. 首位有1到9可选, 二位有0到9除了一位已选的数字可选.

f(3) = f(2)*8. 三位有0到9除了一位二位已选的数字可选.

...

f(10) = f(9)*1.

f(11) = 0 = f(12) = f(13).

最后res是f(1) + f(2) + f(3) + ... + f(n).

Time Complexity: O(n). Space: O(1).

AC Java:

class Solution {
    public int countNumbersWithUniqueDigits(int n) {
        if(n == 0){
            return 1;
        }
        
        int res = 10;
        int uniqueCount = 9;
        int availableNum = 9;
        while(n>1 && availableNum>0){
            uniqueCount *= availableNum;
            res += uniqueCount;
            availableNum--;
            n--;
        }
        return res;
    }
}