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:
- 使用本文中讨论的方法计算数字 N 中的总位数。
- 计算 N 中不同数字的总数。
- 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 <bits/stdc++.h>
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 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 < 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
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 < 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
java 描述语言
<script>
// Javascript program to find next consecutive
// Number with all distinct digits
let INT_MAX = Number.MAX_VALUE;
// Function to count distinct
// digits in a number
function countDistinct(n)
{
// To count the occurrence of digits
// in number from 0 to 9
let arr = new Array(10);
arr.fill(0);
let count = 0;
// Iterate over the digits of the number
// Flag those digits as found in the array
while (n != 0)
{
let r = n % 10;
arr[r] = 1;
n = parseInt(n / 10, 10);
}
// Traverse the array arr and count the
// distinct digits in the array
for (let i = 0; i < 10; i++)
{
if (arr[i] != 0)
count++;
}
return count;
}
// Function to return the total number
// of digits in the number
function countDigit(n)
{
let c = 0;
// Iterate over the digits of the number
while (n != 0)
{
let r = n % 10;
c++;
n = parseInt(n / 10, 10);
}
return c;
}
// Function to return the next
// number with distinct digits
function nextNumberDistinctDigit(n)
{
while (n < INT_MAX)
{
// Count the distinct digits in N + 1
let distinct_digits = countDistinct(n + 1);
// Count the total number of digits in N + 1
let 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;
}
let n = 2019;
document.write(nextNumberDistinctDigit(n));
</script>
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<int> 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 <bits/stdc++.h>
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<int> 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;
}
Output
2031
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