C++ 如何使用递归检查数字是否有重复数字？

我需要知道一个数字是否有使用递归的重复数字，并返回“是”或“否”。我不允许使用循环或数组。这是我到现在为止对10个全局变量所做的，它是有效的，但我认为有更好的方法

#include <iostream>

using namespace std;


int counter0 = 0;
int counter1 = 0;
int counter2 = 0;
int counter3 = 0;
int counter4 = 0;
int counter5 = 0;
int counter6 = 0;
int counter7 = 0;
int counter8 = 0;
int counter9 = 0;
bool check(int k) {
    int p = k % 10;;
    if (k < 10) {
        return false;
    } else {
        if (p == 0) {
            counter0++;
        } else if (p == 1) {
            counter1++;
        } else if (p == 2) {
            counter2++;
        } else if (p == 3) {
            counter3++;
        } else if (p == 4) {
            counter4++;
        } else if (p == 5) {
            counter5++;
        } else if (p == 6) {
            counter6++;
        } else if (p == 7) {
            counter7++;
        } else if (p == 8) {
            counter8++;
        } else if (p == 9) {
            counter9++;
        }
        if(counter1>1 || counter2>1 || counter3>1 || counter4>1 || counter5>1 || counter6>1 || counter7>1 || counter8>1 || counter9>1)
        {
            return true;
        }
        k=k/10;
        check(k);
    }

}

int main() {
    //cout << "Hello, World!" << std::endl;
    int n;
    cin >> n;
   cout << (check(n) ? "yes" : "no") << endl;
    //cout << n/10;
    return 0;
}

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递归问题总是有一个基本情况和一个递归情况

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基本情况很简单：首先，代码不正确。。。 如果输入n=11，它会显示“否”，但1会重复两次。您可以通过将if语句从ifk<10更改为ifk==0来修复它 你可以进入比特级别，但我看不出有多少有用。。。 总之，这是在没有阵列的情况下所能做的最好的

---

但是：如果您需要查找一个数字是否连续重复两次或两次以上，另一个答案是完美的

您当前的代码与556或55或类似的数字不兼容，则不会给出答案。首先，确保您当前的代码工作正常

bool check(int k) {
  if (k < 11)
    return false;
  int digit = k % 10;
  int next = k / 10;
  int digit2 = next % 10;
  if (digit == digit2)
    return true;
  else
    return check(next);
  // Or in one expression:
  // return (digit == digit2) || check(next);
}

#include <iostream>

using namespace std;

bool hasRepeatingDigit(int n, int mask)
{
    // base case: if we have checked all the digits and didn't find any duplicates, return false
    if (n == 0)
        return false;

    /*
        p is the place of the last digit in n (n%10).
        A digit can range from 0 to 9.
        The place of 0 will be 1 << 0 which is 1.
        The place of 1 will be 1 << 1 which is 2.
        The place of 2 will be 1 << 2 which is 4.
        ...
        ...
        The place of 9 will be 1 << 9 which is 512.
    */
    int p = 1 << (n % 10);

    // if place of p has already been marked then it's a duplicate
    if (mask&p)
        return true;

    // otherwise scrap the last digit (n/10), mark place p and recurse on the remaining digits
    return hasRepeatingDigit(n / 10, mask|p);
}

int main()
{
    int n;
    cin >> n;

    cout << hasRepeatingDigit(n, 0) << endl;
}