Given a non-negative integer n, count all numbers with unique digits, x, where 0 ≤ x < 10n.
Example:
Input: 2 Output: 91 Explanation: The answer should be the total numbers in the range of 0 ≤ x < 100, excluding11,22,33,44,55,66,77,88,99
class Solution { public int countNumbersWithUniqueDigits(int n) { if (n == 0) { return 1; } int ret = 10, count = 9; for (int i = 2; i <= n; i++) { count *= 9-i+2; ret += count; } return ret; } }
当n=1时因为只有一个数字，所以0-9都是答案．当n>=2时，最高位可以为1-9任意一个数字，之后各位可以选择的数字个数依次为9, 8, 7, 6...，上一位选一个下一位就少了一种选择．
357. Count Numbers with Unique Digits
原文：https://www.cnblogs.com/wentiliangkaihua/p/11870954.html