Question

Formatted question description: https://leetcode.ca/all/357.html

 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:

Input: 2
Output: 91
Explanation: The answer should be the total numbers in the range of 0 ≤ x < 100,
excluding 11,22,33,44,55,66,77,88,99

@tag-dp
@tag-math


Algorithm

Three digits

11 can be 110 or 101
22 can be 220 or 202
So there are only 8 possibilities

10 +
90-9 +
(90-9) * 8


In four digits

111 can be 1010, 1101, 1110


There are only 7 possibilities

• One-digit number that meets the requirements is 10 (0 to 9)
• The two-digit number that satisfies the meaning of the question is 81, [10-99] remove the [11,22,33,44,55,66,77,88,99] of these 90 numbers, and there is still 81.

The general term formula is f(k) = 9 * 9 * 8 * ... (9-k + 2), then we can pass the [1, n] interval digits through the general term formula according to the size of n Calculate and add up

Java