给定自然数N ，任务是查找与给定数字具有不同数字的下一个数字。
例子：
Input: N = 19
Output: 20
Explanation:
Next number to 19 whose digits are different from 19 is 20.
Input: N = 2019
Output: 3333
Explanation:
Next number to 2019 whose digits are different from 2019 is 3333.
方法：想法是使用散列来计算下一个与给定数字具有不同数字的数字–
- 创建一个哈希数组以存储数字中存在的数字，存储最高有效数字，还将数字中存在的数字存储在count变量中
- 查找数字中不存在的，比当前最高有效位更大的最高有效位的下一个数字。
- 如果未找到下一个最高有效位，则通过增加计数来增加位数，并找到数字中不存在的1到9之间的最高有效位。
- 如果找到下一个最高有效位数，则找到下一个最小位数，该位数在0到9之间的数字中不存在。
- 从1迭代数字以计数
- 将最高有效数字乘以10，然后加上给定数字中不存在的下一个数字。
下面是上述方法的实现：
C++
// C++ implementation to find
// the next distinct digits number
#include
using namespace std;
// Function to find the
// next distinct digits number
void findNextNumber(int n)
{
int h[10] = { 0 };
int i = 0, msb = n, rem = 0;
int next_num = -1, count = 0;
// Loop to find the distinct
// digits using hash array
// and the number of digits
while (msb > 9) {
rem = msb % 10;
h[rem] = 1;
msb /= 10;
count++;
}
h[msb] = 1;
count++;
// Loop to find the most significant
// distinct digit of the next number
for (i = msb + 1; i < 10; i++) {
if (h[i] == 0) {
next_num = i;
break;
}
}
// Condition to check if the number
// is possible with the same number
// of digits count
if (next_num == -1){
for (i = 1; i < msb; i++){
if (h[i] == 0){
next_num = i;
count++;
break;
}
}
}
// Condition to check if the desired
// most siginificant distinct
// digit is found
if (next_num > 0){
// Loop to find the minimum next digit
// which is not present in the number
for (i = 0; i < 10; i++) {
if (h[i] == 0) {
msb = i;
break;
}
}
// Computation of the number
for (i = 1; i < count; i++) {
next_num = ((next_num * 10) + msb);
}
// Condition to check if the number is
// greater than the given number
if (next_num > n)
cout << next_num << "\n";
else
cout << "Not Possible \n";
}
else{
cout << "Not Possible \n";
}
}
// Driver Code
int main()
{
int n = 2019;
findNextNumber(n);
return 0;
} Java
// Java implementation to find
// the next distinct digits number
class GFG{
// Function to find the
// next distinct digits number
static void findNextNumber(int n)
{
int h[] = new int[10];
int i = 0, msb = n, rem = 0;
int next_num = -1, count = 0;
// Loop to find the distinct
// digits using hash array
// and the number of digits
while (msb > 9) {
rem = msb % 10;
h[rem] = 1;
msb /= 10;
count++;
}
h[msb] = 1;
count++;
// Loop to find the most significant
// distinct digit of the next number
for (i = msb + 1; i < 10; i++) {
if (h[i] == 0) {
next_num = i;
break;
}
}
// Condition to check if the number
// is possible with the same number
// of digits count
if (next_num == -1){
for (i = 1; i < msb; i++){
if (h[i] == 0){
next_num = i;
count++;
break;
}
}
}
// Condition to check if the desired
// most siginificant distinct
// digit is found
if (next_num > 0){
// Loop to find the minimum next digit
// which is not present in the number
for (i = 0; i < 10; i++) {
if (h[i] == 0) {
msb = i;
break;
}
}
// Computation of the number
for (i = 1; i < count; i++) {
next_num = ((next_num * 10) + msb);
}
// Condition to check if the number is
// greater than the given number
if (next_num > n)
System.out.print(next_num+ "\n");
else
System.out.print("Not Possible \n");
}
else{
System.out.print("Not Possible \n");
}
}
// Driver Code
public static void main(String[] args)
{
int n = 2019;
findNextNumber(n);
}
}
// This code is contributed by 29AjayKumarC#
// C# implementation to find
// the next distinct digits number
using System;
class GFG{
// Function to find the
// next distinct digits number
static void findNextNumber(int n)
{
int []h = new int[10];
int i = 0, msb = n, rem = 0;
int next_num = -1, count = 0;
// Loop to find the distinct
// digits using hash array
// and the number of digits
while (msb > 9) {
rem = msb % 10;
h[rem] = 1;
msb /= 10;
count++;
}
h[msb] = 1;
count++;
// Loop to find the most significant
// distinct digit of the next number
for (i = msb + 1; i < 10; i++) {
if (h[i] == 0) {
next_num = i;
break;
}
}
// Condition to check if the number
// is possible with the same number
// of digits count
if (next_num == -1){
for (i = 1; i < msb; i++){
if (h[i] == 0){
next_num = i;
count++;
break;
}
}
}
// Condition to check if the desired
// most siginificant distinct
// digit is found
if (next_num > 0){
// Loop to find the minimum next digit
// which is not present in the number
for (i = 0; i < 10; i++) {
if (h[i] == 0) {
msb = i;
break;
}
}
// Computation of the number
for (i = 1; i < count; i++) {
next_num = ((next_num * 10) + msb);
}
// Condition to check if the number is
// greater than the given number
if (next_num > n)
Console.WriteLine(next_num);
else
Console.WriteLine("Not Possible");
}
else{
Console.WriteLine("Not Possible");
}
}
// Driver Code
public static void Main(string[] args)
{
int n = 2019;
findNextNumber(n);
}
}
// This code is contributed by Yash_RPython 3
# Python 3 implementation to find
# the next distinct digits number
# Function to find the
# next distinct digits number
def findNextNumber(n):
h = [0 for i in range(10)]
i = 0
msb = n
rem = 0
next_num = -1
count = 0
# Loop to find the distinct
# digits using hash array
# and the number of digits
while (msb > 9):
rem = msb % 10
h[rem] = 1
msb //= 10
count += 1
h[msb] = 1
count += 1
# Loop to find the most significant
# distinct digit of the next number
for i in range(msb + 1,10,1):
if (h[i] == 0):
next_num = i
break
# Condition to check if the number
# is possible with the same number
# of digits count
if (next_num == -1):
for i in range(1,msb,1):
if (h[i] == 0):
next_num = i
count += 1
break
# Condition to check if the desired
# most siginificant distinct
# digit is found
if (next_num > 0):
# Loop to find the minimum next digit
# which is not present in the number
for i in range(0,10,1):
if (h[i] == 0):
msb = i
break
# Computation of the number
for i in range(1,count,1):
next_num = ((next_num * 10) + msb)
# Condition to check if the number is
# greater than the given number
if (next_num > n):
print(next_num)
else:
print("Not Possible")
else:
print("Not Possible")
# Driver Code
if __name__ == '__main__':
n = 2019
findNextNumber(n)
# This code is contributed by Surendra_Gangwar输出：
3333