Minimum non-"repeating number" (written question)

Topic:

Give a definition: For an integer, if there are two adjacent digits on the same number, then it is called "duplicate number"; now given a positive integer n, the smallest non-"duplicate number" is not less than N.

Ideas:

Assuming that the number of inputs is n, it makes m=n,

Take the minimum two digits of M, B, respectively,

Judge whether the a==b, if the description is the number of repetitions, then recursive call n=m+1; Consider the special case, ab=99, the yield after 100 is still not repeat number, at this time should be recursive call n=m+2;

If a!=b, then move forward one, that is, M=M/10, until m/10=0, and finally return N.

This idea can also be achieved by non-recursive return, see the code.

Code:

#include <iostream>using namespace Std;int calnonrepetitionnum (int n) {if (n<10) return n+1;    int A, B;    int base=1;    int m=n;        while (M/10) {b=m%10;        a=m/10%10;        base*=10;            if (a==b) {if (a==9) return Calnonrepetitionnum ((m+2) *BASE/10);        else return Calnonrepetitionnum ((m+1) *BASE/10);    } m/=10; } return n;}    int calnonrepetitionnum_2 (int n) {if (n<10) return n+1;    int a,b,m;    int base;    BOOL Flag=true;        while (flag) {flag=false;        M=n;        base=1;            while (M/10) {a=m%10;            b=m/10%10;            base*=10;                    if (a==b) {if (a==9) {n= (m+2) *BASE/10;                    Flag=true;                Break                    } else{n= (m+1) *BASE/10;                    Flag=true;                Break       }            }     M=M/10; }} return n;}    int main () {cout << calnonrepetitionnum () << Endl;    cout << calnonrepetitionnum_2 () << Endl; return 0;}

　　

Minimum non-"repeating number" (written question)