具有唯一(或不同)数字的数字

给定范围，请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.

Javascript

 
// Javascript implementation to find unique digit
// numbers in a range
 
// Function to print unique digit numbers
// in range from l to r.
function printUnique(l, r)
{
    // Start traversing the numbers
    for (let i=l ; i<=r ; i++)
    {
        let num = i;
        let visited = new Array(10);
 
        // 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 = Math.floor(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)
            document.write(i + " ");
    }
}
 
// Driver code
 
    let l = 1, r = 20;
    printUnique(l, r);
     
// This code is contributed by Mayank Tyagi
 

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为了检查一个数字是否只有唯一数字。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。

这是上述方法的代码：

C++

// CPP 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;
}

输出

1 2 3 4 5 6 7 8 9 10 12 13 14 15 16 17 18 19 20