很典型的數位dp，把全球第一的代碼拿過來研究了一下，加了點註釋

代碼作者：waakaaka

個人主頁：https://leetcode.com/waakaaka/

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 #define db(x) cerr << #x << "=" << x << endl
 4 #define db2(x, y) cerr << #x << "=" << x << "," << #y << "=" << y << endl
 5
 
 #define db3(x, y, z) cerr << #x << "=" << x << "," << #y << "=" << y << "," << #z << "=" << z << endl
 6 #define dbv(v) cerr << #v << "="; for (auto _x : v) cerr << _x << ", "; cerr << endl
 7 #define
 
 dba(a, n) cerr << #a << "="; for (int _i = 0; _i < (n); ++_i) cerr << a[_i] << ", "; cerr << endl
 8 typedef long long ll;
 9 typedef long double ld;
10 class Solution
11 {
12     public:
13         int n;
14         int dis;
15         //val為當前數值，bs可以看作是vector<int> hash(10)
 

16         void go(ll val, int bs)
17         {
18             // 當前數值小於題目所給n，非重復的結果數++
19             if (val <= n) ++dis;
20             //當前數值乘以10大於所給n，直接返回，不然進了下面的循環就乘以10繼續遞歸了
21             if (val * 10 > n) return;
22             for (int i = 0; i <= 9; ++i)
23             {
24                 //當前數值為0（bs為零等價於val為零）且加的一個數字還是零，就直接continue，因為不支持前綴零
25                 if (!bs && i == 0) continue; // no 0 for first digit
26                 //如果要加的數字i和已經有的數字重復，就也continue，不進行遞歸 
27                 if (bs & (1 << i)) continue;
28                 //加上尾數i，然後更新狀態bs，繼續遞歸 
29                 go(val * 10 + i, bs | (1 << i));
30             }
31         }
32         int numDupDigitsAtMostN(int N)
33         {
34             dis = 0;
35             n = N;
36             go(0, 0);
37             //dis為不重復的數字的個數 
38             return N + 1 - dis;
39         }
40 };

Leetcode-1015 Numbers With Repeated Digits(至少有 1 位重復的數字)