9. Count the number of numbers with different numbers
9.1. Requirements for the title
Given an integer n, count and return the number of numbers x with different digits, where 0 <= x < 10n.
示例 1：
输入：n = 2
输出：91
解释：答案应为除去 11、22、33、44、55、66、77、88、99 外，在 0 ≤ x < 100 范围内的所有数字。
示例 2：
输入：n = 0
输出：1
提示：
0 <= n <= 8
Source: LeetCode
Link: https://leetcode-cn.com/problems/count-numbers-with-unique-digits
9.2, problem-solving ideas
the first method:
When n=0, the value is only one, which is not equal;
when n=1, the value is [0,9], and the 10-digit numbers are not equal;
when n>1, there is a formula 
, which can be calculated according to the formula calculate.
The second method:
Use the hit meter directly.
9.3. Algorithms
the first method
class Solution {
public int countNumbersWithUniqueDigits(int n) {
//n=0
if(n == 0){
return 1;
}
//n=1
if(n == 1){
return 10;
}
//n>1
int count = 9,ans = 10,a = 9;
while(n > 1 && count > 0){
a *= count;
ans += a;
count--;
n--;
}
return ans;
}
}
The second method
class Solution {
public int countNumbersWithUniqueDigits(int n) {
int [] array = {
1,10,91,739,5275,32491,168571,712891,2345851};
return array[n];
}
}