具有不同数字的下一个数字

给定整数N ，任务是查找其中具有不同数字的下一个数字。
例子：

Input: N = 20 
Output: 21 
The next integer with all distinct digits after 20 is 21.
Input: N = 2019 
Output: 2031 

方法：

1. 使用本文讨论的方法计算数字N中的位数总数。
2. 计算N中不同位数的总数。
3. 如果N中的总位数与不同位数的数量相等，则返回该数字，否则将数字加1并重复前面的步骤。

下面是上述方法的实现：

C++

// C++ program to find next consecutive
// Number with all distinct digits
#include 
using namespace std;
 
// Function to count distinct
// digits in a number
int countDistinct(int n)
{
    // To count the occurrence of digits
    // in number from 0 to 9
    int arr[10] = { 0 };
    int count = 0;
 
    // Iterate over the digits of the number
    // Flag those digits as found in the array
    while (n) {
        int r = n % 10;
        arr[r] = 1;
        n /= 10;
    }
 
    // Traverse the array arr and count the
    // distinct digits in the array
    for (int i = 0; i < 10; i++) {
        if (arr[i])
            count++;
    }
    return count;
}
 
// Function to return the total number
// of digits in the number
int countDigit(int n)
{
    int c = 0;
 
    // Iterate over the digits of the number
    while (n) {
        int r = n % 10;
        c++;
        n /= 10;
    }
    return c;
}
 
// Function to return the next
// number with distinct digits
int nextNumberDistinctDigit(int n)
{
    while (n < INT_MAX) {
 
        // Count the distinct digits in N + 1
        int distinct_digits = countDistinct(n + 1);
 
        // Count the total number of digits in N + 1
        int total_digits = countDigit(n + 1);
 
        if (distinct_digits == total_digits) {
 
            // Return the next consecutive number
            return n + 1;
        }
 
        else
            // Increment Number by 1
            n++;
    }
    return -1;
}
 
// Driver code
int main()
{
    int n = 2019;
 
    cout << nextNumberDistinctDigit(n);
 
    return 0;
}

Java

// Java program to find next consecutive
// Number with all distinct digits
class GFG
{
     
    final static int INT_MAX = Integer.MAX_VALUE ;
     
    // Function to count distinct
    // digits in a number
    static int countDistinct(int n)
    {
         
        // To count the occurrence of digits
        // in number from 0 to 9
        int arr[] = new int[10];
        int count = 0;
     
        // Iterate over the digits of the number
        // Flag those digits as found in the array
        while (n != 0)
        {
            int r = n % 10;
            arr[r] = 1;
            n /= 10;
        }
     
        // Traverse the array arr and count the
        // distinct digits in the array
        for (int i = 0; i < 10; i++)
        {
            if (arr[i] != 0)
                count++;
        }
        return count;
    }
     
    // Function to return the total number
    // of digits in the number
    static int countDigit(int n)
    {
        int c = 0;
     
        // Iterate over the digits of the number
        while (n != 0)
        {
            int r = n % 10;
            c++;
            n /= 10;
        }
        return c;
    }
     
    // Function to return the next
    // number with distinct digits
    static int nextNumberDistinctDigit(int n)
    {
        while (n < INT_MAX)
        {
     
            // Count the distinct digits in N + 1
            int distinct_digits = countDistinct(n + 1);
     
            // Count the total number of digits in N + 1
            int total_digits = countDigit(n + 1);
     
            if (distinct_digits == total_digits)
            {
     
                // Return the next consecutive number
                return n + 1;
            }
     
            else
             
                // Increment Number by 1
                n++;
        }
        return -1;
    }
     
    // Driver code
    public static void main (String[] args)
    {
        int n = 2019;
     
        System.out.println(nextNumberDistinctDigit(n));
    }
}
 
// This code is contributed by AnkitRai01

Python3

# Python3 program to find next consecutive
# Number with all distinct digits
import sys
 
INT_MAX = sys.maxsize;
 
# Function to count distinct
# digits in a number
def countDistinct(n):
 
    # To count the occurrence of digits
    # in number from 0 to 9
    arr = [0] * 10;
    count = 0;
 
    # Iterate over the digits of the number
    # Flag those digits as found in the array
    while (n != 0):
        r = int(n % 10);
        arr[r] = 1;
        n //= 10;
     
    # Traverse the array arr and count the
    # distinct digits in the array
    for i in range(10):
        if (arr[i] != 0):
            count += 1;
     
    return count;
 
# Function to return the total number
# of digits in the number
def countDigit(n):
    c = 0;
 
    # Iterate over the digits of the number
    while (n != 0):
        r = n % 10;
        c+=1;
        n //= 10;
     
    return c;
 
# Function to return the next
# number with distinct digits
def nextNumberDistinctDigit(n):
    while (n < INT_MAX):
 
        # Count the distinct digits in N + 1
        distinct_digits = countDistinct(n + 1);
 
        # Count the total number of digits in N + 1
        total_digits = countDigit(n + 1);
 
        if (distinct_digits == total_digits):
 
            # Return the next consecutive number
            return n + 1;
        else:
 
            # Increment Number by 1
            n += 1;
     
    return -1;
 
# Driver code
if __name__ == '__main__':
    n = 2019;
 
    print(nextNumberDistinctDigit(n));
     
# This code is contributed by PrinciRaj1992

C#

// C# program to find next consecutive
// Number with all distinct digits
using System;
 
class GFG
{
     
    readonly static int INT_MAX = int.MaxValue ;
     
    // Function to count distinct
    // digits in a number
    static int countDistinct(int n)
    {
         
        // To count the occurrence of digits
        // in number from 0 to 9
        int []arr = new int[10];
        int count = 0;
     
        // Iterate over the digits of the number
        // Flag those digits as found in the array
        while (n != 0)
        {
            int r = n % 10;
            arr[r] = 1;
            n /= 10;
        }
     
        // Traverse the array arr and count the
        // distinct digits in the array
        for (int i = 0; i < 10; i++)
        {
            if (arr[i] != 0)
                count++;
        }
        return count;
    }
     
    // Function to return the total number
    // of digits in the number
    static int countDigit(int n)
    {
        int c = 0;
     
        // Iterate over the digits of the number
        while (n != 0)
        {
            int r = n % 10;
            c++;
            n /= 10;
        }
        return c;
    }
     
    // Function to return the next
    // number with distinct digits
    static int nextNumberDistinctDigit(int n)
    {
        while (n < INT_MAX)
        {
     
            // Count the distinct digits in N + 1
            int distinct_digits = countDistinct(n + 1);
     
            // Count the total number of digits in N + 1
            int total_digits = countDigit(n + 1);
     
            if (distinct_digits == total_digits)
            {
     
                // Return the next consecutive number
                return n + 1;
            }
     
            else
             
                // Increment Number by 1
                n++;
        }
        return -1;
    }
     
    // Driver code
    public static void Main(String[] args)
    {
        int n = 2019;
     
        Console.WriteLine(nextNumberDistinctDigit(n));
    }
}
 
// This code is contributed by PrinciRaj1992

输出

2031

另一种方法：

不必每次都计算数字位数，而是可以使用set STL来检查数字是否只有唯一的数字。

然后，我们可以比较由给定数字和新创建的集合形成的字符串s的大小。

例如，让我们考虑数字1987，然后我们可以将数字转换为字符串，

C++

int n;
cin>>n;
string s = to_string(n);

之后，使用字符串s的内容初始化一个集合。

C++

set uniDigits(s.begin(), s.end());

然后，我们可以比较字符串s和新创建的集合uniDigits的大小。

这是总代码

C++

// CPP program for the above program
#include 
using namespace std;
 
// Function to find next number
// with digit distinct
void nextNumberDistinctDigit(int n)
{
     
    // Iterate from n + 1 to inf
    for (int i = n + 1;; i++) {
       
        // Convert the no. to
        // string
        string s = to_string(i);
       
        // Convert string to set using stl
        set uniDigits(s.begin(), s.end());
       
        // Output if condition satisfies
        if (s.size() == uniDigits.size()) {
            cout << i;
            break;
        }
    }
}
 
// Driver Code
int main()
{
    int n = 2019; // input the no.
     
    // Function Call
    nextNumberDistinctDigit(n);
    return 0;
}

输出

2031