[Leetcode] 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 is the total numbers in the range of 0≤x <, excluding [11,22,33,44,55,66,77,88,99])

Hint:

• A Direct backtracking approach.
• Backtracking should contains three states which is (the current number, number of steps to get, and a bitmask which represent which number is marked as visited so far in the current number). Start with State (0,0,0) and count all valid number till we reach number of steps equals to 10n.
• This problem can also is solved using a dynamic programming approach and some knowledge of combinatorics.
• Let f (k) = Count of numbers with the unique digits with length equals K.
• F (1) = ten, ..., f (k) = 9 * 9 * 8 * ... (9-k + 2) [The first factor is 9 because a number cannot start with 0].

Thinking of solving problems

Calculated as the formula in the prompt.

Implementation code

//runtime:0 MS Public  class solution {     Public int countnumberswithuniquedigits(intN) {if(n = =0) {return 1; }intTotal =Ten;intCNT =9; for(inti =2; I <= N; i++) {CNT *= OneI        Total + = cnt; }returnTotal }}

[Leetcode] Count Numbers with Unique Digits