给定一个范围,打印所有具有唯一数字的数字。
例子 :
Input : 10 20
Output : 10 12 13 14 15 16 17 18 19 20 (Except 11)
Input : 1 10
Output : 1 2 3 4 5 6 7 8 9 10
方法:
As the problem is pretty simple, the only thing to be done is :-
1- Find the digits one by one and keep marking visited digits.
2- If all digits occurs one time only then print that number.
3- Else not.
C++
// C++ implementation to find unique digit
// numbers in a range
#include
using namespace std;
// Function to print unique digit numbers
// in range from l to r.
void printUnique(int l, int r)
{
// Start traversing the numbers
for (int i=l ; i<=r ; i++)
{
int num = i;
bool visited[10] = {false};
// Find digits and maintain its hash
while (num)
{
// if a digit occurs more than 1 time
// then break
if (visited[num % 10])
break;
visited[num%10] = true;
num = num/10;
}
// num will be 0 only when above loop
// doesn't get break that means the
// number is unique so print it.
if (num == 0)
cout << i << " ";
}
}
// Driver code
int main()
{
int l = 1, r = 20;
printUnique(l, r);
return 0;
}
Java
// Java implementation to find unique digit
// numbers in a range
class Test
{
// Method to print unique digit numbers
// in range from l to r.
static void printUnique(int l, int r)
{
// Start traversing the numbers
for (int i=l ; i<=r ; i++)
{
int num = i;
boolean visited[] = new boolean[10];
// Find digits and maintain its hash
while (num != 0)
{
// if a digit occurs more than 1 time
// then break
if (visited[num % 10])
break;
visited[num%10] = true;
num = num/10;
}
// num will be 0 only when above loop
// doesn't get break that means the
// number is unique so print it.
if (num == 0)
System.out.print(i + " ");
}
}
// Driver method
public static void main(String args[])
{
int l = 1, r = 20;
printUnique(l, r);
}
}
Python3
# Python3 implementation
# to find unique digit
# numbers in a range
# Function to print
# unique digit numbers
# in range from l to r.
def printUnique(l,r):
# Start traversing
# the numbers
for i in range (l, r + 1):
num = i;
visited = [0,0,0,0,0,0,0,0,0,0];
# Find digits and
# maintain its hash
while (num):
# if a digit occurs
# more than 1 time
# then break
if visited[num % 10] == 1:
break;
visited[num % 10] = 1;
num = (int)(num / 10);
# num will be 0 only when
# above loop doesn't get
# break that means the
# number is unique so
# print it.
if num == 0:
print(i, end = " ");
# Driver code
l = 1;
r = 20;
printUnique(l, r);
# This code is
# contributed by mits
C#
// C# implementation to find unique digit
// numbers in a range
using System;
public class GFG {
// Method to print unique digit numbers
// in range from l to r.
static void printUnique(int l, int r)
{
// Start traversing the numbers
for (int i = l ; i <= r ; i++)
{
int num = i;
bool []visited = new bool[10];
// Find digits and maintain
// its hash
while (num != 0)
{
// if a digit occurs more
// than 1 time then break
if (visited[num % 10])
break;
visited[num % 10] = true;
num = num / 10;
}
// num will be 0 only when
// above loop doesn't get
// break that means the number
// is unique so print it.
if (num == 0)
Console.Write(i + " ");
}
}
// Driver method
public static void Main()
{
int l = 1, r = 20;
printUnique(l, r);
}
}
// This code is contributed by Sam007.
PHP
Javascript
C++
int n;
cin>>n;
string s = to_string(n);
C++
set uniDigits(s.begin(), s.end());
C++
// C++ code for the above approach
#include
using namespace std;
// Function to print unique
// numbers
void printUnique(int l, int r){
// Iterate from l to r
for (int i = l; i <= r; 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 << " ";
}
}
}
// Driver Code
int main()
{
// Input of the lower and
// higher limits
int l = 1, r = 20;
// Function Call
printUnique(l, r);
return 0;
}
Java
// Java code for the above approach
import java.util.*;
class GFG{
// Function to print unique
// numbers
static void printUnique(int l, int r)
{
// Iterate from l to r
for(int i = l; i <= r; i++)
{
// Convert the no. to
// String
String s = String.valueOf(i);
// Convert String to set using stl
HashSet uniDigits = new HashSet();
for(int c : s.toCharArray())
uniDigits.add(c);
// Output if condition satisfies
if (s.length() == uniDigits.size())
{
System.out.print(i+ " ");
}
}
}
// Driver Code
public static void main(String[] args)
{
// Input of the lower and
// higher limits
int l = 1, r = 20;
// Function Call
printUnique(l, r);
}
}
// This code is contributed by Princi Singh
C#
// C# code for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to print unique
// numbers
static void printUnique(int l, int r)
{
// Iterate from l to r
for(int i = l; i <= r; i++)
{
// Convert the no. to
// String
String s = String.Join("", i);
// Convert String to set using stl
HashSet uniDigits = new HashSet();
foreach(int c in s.ToCharArray())
uniDigits.Add(c);
// Output if condition satisfies
if (s.Length == uniDigits.Count)
{
Console.Write(i + " ");
}
}
}
// Driver Code
public static void Main(String[] args)
{
// Input of the lower and
// higher limits
int l = 1, r = 20;
// Function Call
printUnique(l, r);
}
}
// This code is contributed by Princi Singh
Javascript
输出:
1 2 3 4 5 6 7 8 9 10 12 13 14 15 16 17 18 19 20
另一种方法:
使用 set STL 来检查一个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,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。
以下是上述方法的代码:
C++
// C++ code for the above approach
#include
using namespace std;
// Function to print unique
// numbers
void printUnique(int l, int r){
// Iterate from l to r
for (int i = l; i <= r; 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 << " ";
}
}
}
// Driver Code
int main()
{
// Input of the lower and
// higher limits
int l = 1, r = 20;
// Function Call
printUnique(l, r);
return 0;
}
Java
// Java code for the above approach
import java.util.*;
class GFG{
// Function to print unique
// numbers
static void printUnique(int l, int r)
{
// Iterate from l to r
for(int i = l; i <= r; i++)
{
// Convert the no. to
// String
String s = String.valueOf(i);
// Convert String to set using stl
HashSet uniDigits = new HashSet();
for(int c : s.toCharArray())
uniDigits.add(c);
// Output if condition satisfies
if (s.length() == uniDigits.size())
{
System.out.print(i+ " ");
}
}
}
// Driver Code
public static void main(String[] args)
{
// Input of the lower and
// higher limits
int l = 1, r = 20;
// Function Call
printUnique(l, r);
}
}
// This code is contributed by Princi Singh
C#
// C# code for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to print unique
// numbers
static void printUnique(int l, int r)
{
// Iterate from l to r
for(int i = l; i <= r; i++)
{
// Convert the no. to
// String
String s = String.Join("", i);
// Convert String to set using stl
HashSet uniDigits = new HashSet();
foreach(int c in s.ToCharArray())
uniDigits.Add(c);
// Output if condition satisfies
if (s.Length == uniDigits.Count)
{
Console.Write(i + " ");
}
}
}
// Driver Code
public static void Main(String[] args)
{
// Input of the lower and
// higher limits
int l = 1, r = 20;
// Function Call
printUnique(l, r);
}
}
// This code is contributed by Princi Singh
Javascript
输出
1 2 3 4 5 6 7 8 9 10 12 13 14 15 16 17 18 19 20