357. Count Numbers with Unique Digits

题目链接：https://leetcode.com/problems/count-numbers-with-unique-digits/

题目描述：

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 be the total numbers in the range of 0 ≤ x < 100, excluding [11,22,33,44,55,66,77,88,99])

Hint:

1. A direct way is to use the backtracking approach.
2. Backtracking should contains three states which are (the current number, number of steps to get that number 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.
3. This problem can also be solved using a dynamic programming approach and some knowledge of combinatorics.
4. Let f(k) = count of numbers with unique digits with length equals k.
5. f(1) = 10, ..., f(k) = 9 * 9 * 8 * ... (9 - k + 2) [The first factor is 9 because a number cannot start with 0].

思路一：回溯法（超时）

class Solution {public:int countNumbersWithUniqueDigits(int n) {int cnt = 0;if(n==0) cnt=1;map<int, int> bitmask;int maxstep = pow(10, n);dfs("", 0, bitmask, cnt, maxstep, n);return cnt;}void dfs(string str, int step, map<int, int> bitmask, int& cnt, int maxstep, int n){if (step == maxstep || str.size() == n)return;for (int i = 0; i<10; i++){if (str == "0") break;step++;if (bitmask[i] == 1)continue;else{bitmask[i] = 1;string str1 = str + to_string(i);cnt++;dfs(str1, step, bitmask, cnt, maxstep, n);}bitmask[i] = 0;}}};

思路二：动态规划，用排列组合

class Solution {public:int countNumbersWithUniqueDigits(int n) {if (n == 0||n==1) return pow(10,n);    int fn=10;    int tmp=9;for(int i=2;i<=n;i++){    tmp*=10-i+1;    fn=fn+tmp;}return fn;}};