[leetcode]357. Count Numbers with Unique Digits

题目链接:https://leetcode.com/problems/count-numbers-with-unique-digits/

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：n=0, 返回1

                  n=1,返回1+9

                  n=2,返回1+9+9*9

                  n=3,返回1+9+9*9*8

                  ....

                 n=k,返回1+9+9*9*8+...+9*9*8*...*(11-k)

                 n>10,不再变化

class Solution{
public:
    int countNumbersWithUniqueDigits(int n)
    {
        if(n==0) return 1;
        if(n==1) return 10;
        int count=10,faciend=9;
        for(int i=2;i<=n&&i<=10;i++)
        {
            faciend*=(11-i);
            count+=faciend;
        }
        return count;
    }
};