# "Leetcode" Count Numbers with Unique Digits

Source: Internet
Author: User

Topic:
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].

Ideas:
Recursive equations are given, fools use backtracking. = =: And the backtracking looks so hard.

Algorithm:

``     Public intCountnumberswithuniquedigits (intN) {if(n = =0)return 1;int Count=0; for(inti =1; I <= N; i++) {//How many unique digits each person has            Count+ = cal (i); }return Count; } Public intCalintK) {int sum=9;if(k = =0)return 0;if(k = =1)return Ten; for(inti = OneK I <=9; i++) {sum*= i; }return sum; }``

"Leetcode" Count Numbers with Unique Digits

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