下一个数字不同的数字

原文:https://www . geesforgeks . org/next-number-with-distinct-digits/

给定一个整数 N ，任务是找到下一个数字，其中有不同的数字。

示例:

输入: N = 20
输出:21
20 后所有不同数字的下一个整数是 21。

输入:N = 2019
T3】输出: 2031

进场:

1. 使用本文中讨论的方法计算数字 N 中的总位数。


2. 计算 N 中不同数字的总数。


3. 如果总位数和 N 中不同位数的计数相等，则返回该数，否则将该数增加 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         // 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 <    return 0;
}


Java 语言(一种计算机语言，尤用于创建网站)

// 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         {
            // 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


Python 3

# 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         # 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         {
            // 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


java 描述语言

Output

2031


另一种方法:

我们可以使用 set STL 来检查一个数字是否只有唯一的数字，而不是每次都计算数字的数量。

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的大小。

例如，让我们考虑数字 1987，然后我们可以将该数字转换为字符串，

C++

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


之后，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 的大小和新创建的一组单音数字。

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 <            break;
        }
    }
}
// Driver Code
int main()
{
    int n = 2019; // input the no.
    // Function Call
    nextNumberDistinctDigit(n);
    return 0;
}


Output

2031