Leetcode-1015 Numbers With Repeated Digits(至少有 1 位重复的数字)

很典型的数位dp，把全球第一的代码拿过来研究了一下，加了点注释 

代码作者：waakaaka 

个人主页：https://leetcode.com/waakaaka/

 

#include <bits/stdc++.h>
using namespace std;
#define db(x) cerr << #x << "=" << x << endl
#define db2(x, y) cerr << #x << "=" << x << "," << #y << "=" << y << endl
#define db3(x, y, z) cerr << #x << "=" << x << "," << #y << "=" << y << "," << #z << "=" << z << endl
#define dbv(v) cerr << #v << "="; for (auto _x : v) cerr << _x << ", "; cerr << endl
#define dba(a, n) cerr << #a << "="; for (int _i = 0; _i < (n); ++_i) cerr << a[_i] << ", "; cerr << endl
typedef long long ll;
typedef long double ld;
class Solution
{
    public:
        int n;
        int dis;
        //val为当前数值，bs可以看作是vector<int> hash(10)
        void go(ll val, int bs)
        {
            // 当前数值小于题目所给n，非重复的结果数++
            if (val <= n) ++dis;
            //当前数值乘以10大于所给n，直接返回，不然进了下面的循环就乘以10继续递归了
            if (val * 10 > n) return;
            for (int i = 0; i <= 9; ++i)
            {
                //当前数值为0（bs为零等价于val为零）且加的一个数字还是零，就直接continue，因为不支持前缀零
                if (!bs && i == 0) continue; // no 0 for first digit
                //如果要加的数字i和已经有的数字重复，就也continue，不进行递归 
                if (bs & (1 << i)) continue;
                //加上尾数i，然后更新状态bs，继续递归 
                go(val * 10 + i, bs | (1 << i));
            }
        }
        int numDupDigitsAtMostN(int N)
        {
            dis = 0;
            n = N;
            go(0, 0);
            //dis为不重复的数字的个数 
            return N + 1 - dis;
        }
};