Count Numbers with Unique Digits

Code

Time Complexity: $O(n)$, Space Complexity: $O(1)$

排列組合問題，不重複的排法是階層。

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

Link

• Count Numbers with Unique Digits