leetcode-java-357. Count Numbers with Unique Digits

/*Given a non-negative integer n, count all numbers with unique digits, x, where 0 ≤ x < 10^n.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]) *//*思路:要想数字不一样，用到数学中的排列组合：当n=1时;因为只有一个数字，所以0-9都是答案．当n>=2时;第一位是非0，则有9种情况第二位和第一位不同，也有9种情况第三位有8种情况...... */ public class Solution {     public int countNumbersWithUniqueDigits(int n) {         if(n == 0) {             return 1;         }         if(n == 1) {             return 10;         }         int digits = 9,             val = 10;         for(int i = 2;i <= n;i++) {             digits = digits*(9 - i + 2);             val = val + digits;         }         return val;     } }