leetcode -- 357. Count Numbers with Unique Digits

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,
excluding 11,22,33,44,55,66,77,88,99

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/count-numbers-with-unique-digits
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

我的思路1：（暴力）

1、最直接的就是暴力解，但是超时。到7的时候就超时了

class Solution {
public:int countNumbersWithUniqueDigits(int n) {int len =pow(10,n);int count = 0;for(int i = 0; i < len; i++){if(check(i) == false) count++;}return len - count;}bool check(int n){  //如果有重复，返回falsevector<bool> vis(10, false);if(n == 0) return true;while(n != 0){int yushu = n % 10;if(vis[yushu] == false){vis[yushu] = true;}else return false;n /= 10;}return true;}
};

参考思路2：（dp）

1、dp[i]表示i位数字不同的数字数

class Solution {
public:int countNumbersWithUniqueDigits(int n) {if(n == 0) return 1;int dp[11];dp[0] = 1;int per = 1;for(int i = 1; i <= 10; i++){dp[i] = dp[i-1] + 9*per;per = per * (10-i);}if(n <= 10) return dp[n];else return dp[10];}};

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